Electrostatic Potential and Capacitance: NCERT Solutions

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance are provided here by SimplyAcad to help students get a clear insight of the chapter. The solutions will help students with regular practice and ensure that all their doubts and confusions regarding the chapter vanish away. The students must grab this opportunity to prepare themselves for the upcoming boards examinations.

Electrostatic Potential and Capacitance is an extremely crucial topic for your physics examinations and must be dealt with seriousness to understand the core of the subject. Scroll below to find the NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance.

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Question 1. Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution :Given,

Two charges qA = 5 x 10-8 C and qB = -3×10-8 C

Distance between two charges, r = 16 cm = 0.16 m

Consider a point O on the line joining two charges where the electric potential is zero due to two charges.

Question2. A regular hexagon of side 10 cm has a charge of \( 5 \, \mu\text{C} \) at each of its vertices. Calculate the potential at the center of the hexagon. \section*

Solution- \textbf{1. Geometry of the Hexagon:} For a regular hexagon, the distance from the center to any vertex (the radius of the circumscribed circle) is equal to the side length of the hexagon. Therefore, the distance \( r \) from the center of the hexagon to each vertex is: \[ r = 10 \, \text{cm} = 0.1 \, \text{m} \] \textbf{2. Electric Potential Due to a Single Charge:} The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by: \[ V = \frac{kq}{r} \] where \( k \) is Coulomb’s constant (\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)). \textbf{3. Total Potential at the Center:} Since the hexagon is regular and all charges are the same and equidistant from the center, the potential at the center due to each charge will be the same. The total potential at the center is the sum of the potentials due to each charge. \[ V_{\text{total}} = \sum_{i=1}^6 \frac{kq}{r} \] where there are 6 charges, each contributing the same potential. Substituting the values: \[ V_{\text{total}} = 6 \times \frac{k \times 5 \times 10^{-6}}{0.1} \] \[ V_{\text{total}} = 6 \times \frac{8.99 \times 10^9 \times 5 \times 10^{-6}}{0.1} \] \[ V_{\text{total}} = 6 \times \frac{44.95 \times 10^3}{0.1} \] \[ V_{\text{total}} = 6 \times 449.5 \times 10^3 \] \[ V_{\text{total}} = 2.697 \times 10^6 \, \text{V} \] \textbf{Conclusion:} The electric potential at the center of the hexagon is \( 2.697 \times 10^6 \, \text{V} \).

Question 3. {Question} Two charges, \(2 \, \mu\text{C}\) and \(-2 \, \mu\text{C}\), are placed at points \(A\) and \(B\), which are 6 cm apart. \begin{enumerate} \item[(a)] Identify an equipotential surface of the system. \item[(b)] What is the direction of the electric field at every point on this surface? \end{enumerate} \section*

Solution: \textbf{(a) Equipotential Surface} To find the equipotential surface where the potential is zero everywhere, we need to consider the superposition of the potentials due to the two charges. 1. The potential \(V\) due to a point charge \(q\) at a distance \(r\) is given by: \[ V = \frac{kq}{r} \] where \(k\) is Coulomb’s constant (\(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\)). 2. Let the distance between the charges be \(AB = 6 \, \text{cm} = 0.06 \, \text{m}\). – The charge \(+2 \, \mu\text{C}\) is at point \(A\). – The charge \(-2 \, \mu\text{C}\) is at point \(B\). Consider a point \(P\) on the equipotential surface. Let \(d_1\) be the distance from \(P\) to \(A\) and \(d_2\) be the distance from \(P\) to \(B\). The condition for an equipotential surface is that the total potential due to both charges must be zero: \[ V_A + V_B = 0 \] Substituting the potentials: \[ \frac{k \cdot 2 \times 10^{-6}}{d_1} + \frac{k \cdot (-2 \times 10^{-6})}{d_2} = 0 \] Simplifying: \[ \frac{2}{d_1} = \frac{2}{d_2} \] \[ d_1 = d_2 \] This implies that the equipotential surface is the plane equidistant from both charges, which is at the midpoint of line \(AB\). The plane is perpendicular to line \(AB\). \textbf{(b) Direction of the Electric Field} The electric field direction is always normal to the equipotential surface and points in the direction of decreasing potential. 1. On the equipotential surface (which is the plane equidistant from the two charges), the electric field will be directed towards the line \(AB\) because it points from the positive charge to the negative charge. 2. Direction of the Electric Field: The electric field at every point on this surface will be perpendicular to the surface and will point along the line connecting the two charges \(A\) and \(B\), from the positive charge towards the negative charge.

Question5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = \(10^{-12}\) F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6? \section*

Solution Given: The capacitance of the parallel plate capacitor is \(8 \, \text{pF}\). The capacitance \(C\) of a parallel plate capacitor is given by: \[ C = \frac{k \epsilon_0 A}{d} \] where: \begin{itemize} \item \(C\) is the capacitance, \item \(A\) is the area of each plate, \item \(\epsilon_0\) is the permittivity of free space, \item \(d\) is the distance between the parallel plates, \item \(k\) is the dielectric constant. \end{itemize} If the distance between the plates is reduced by half and the space between them is filled with a dielectric material with a dielectric constant \(k = 6\), the new capacitance \(C’\) can be calculated as follows: \[ C’ = \frac{k \epsilon_0 A}{d/2} \] \[ C’ = \frac{6 \epsilon_0 A}{d/2} = 12 \cdot \frac{\epsilon_0 A}{d} = 12 \cdot C \] Substituting the given value of \(C\): \[ C’ = 12 \cdot 8 \, \text{pF} = 96 \, \text{pF} \] Thus, if the distance between the plates is reduced by half and the space between them is filled with a dielectric material of constant 6, then the capacitance of the capacitor is \(96 \, \text{pF}\).

Question 6. 1) Three capacitors each of capacitance \( 9 \, \text{pF} \) are connected in series. \begin{itemize} \item[(A)] What is the total capacitance of the combination? \end{itemize} 2) Three capacitors each of capacitance \( 9 \, \text{pF} \) are connected in series. \begin{itemize} \item[(B)] What is the potential difference across each capacitor if the combination is connected to a \( 120 \, \text{V} \) supply? \end{itemize} \section*

{Solution} 1) Given, \( C = 9 \, \text{pF} \). For capacitors connected in series, the total capacitance \( C_{\text{eq}} \) is given by: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] Since \( C_1 = C_2 = C_3 = C \), we have: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \] Thus: \[ C_{\text{eq}} = \frac{C}{3} \] Substituting \( C = 9 \, \text{pF} \): \[ C_{\text{eq}} = \frac{9 \, \text{pF}}{3} = 3 \, \text{pF} \] 2) The total potential difference \( V \) across the series combination is \( 120 \, \text{V} \). The charge \( q \) on each capacitor is given by: \[ q = C_{\text{eq}} \cdot V \] Substituting \( C_{\text{eq}} = 3 \, \text{pF} \) and \( V = 120 \, \text{V} \): \[ q = 3 \, \text{pF} \cdot 120 \, \text{V} = 360 \, \text{pC} \] The potential difference across each capacitor \( V’ \) is: \[ V’ = \frac{q}{C} \] Substituting \( q = 360 \, \text{pC} \) and \( C = 9 \, \text{pF} \): \[ V’ = \frac{360 \, \text{pC}}{9 \, \text{pF}} = 40 \, \text{V} \] Thus, the potential difference across each capacitor is \( 40 \, \text{V} \).

Question7. Three capacitors of capacitances \(2 \, \text{pF}\), \(3 \, \text{pF}\), and \(4 \, \text{pF}\) are connected in parallel. \begin{itemize} \item[(a)] What is the total capacitance of the combination? \item[(b)] Determine the charge on each capacitor if the combination is connected to a \(100 \, \text{V}\) supply. \end{itemize} \section*

Solution- \textbf{Part (a)} Step 1: Equivalent capacitance For capacitors connected in parallel, the total capacitance \( C_{\text{eq}} \) is given by: \[ C_{\text{eq}} = C_1 + C_2 + C_3 \] Substituting the given values: \[ C_{\text{eq}} = 2 \, \text{pF} + 3 \, \text{pF} + 4 \, \text{pF} = 9 \, \text{pF} \] \textbf{Part (b)} Step 2: Charge on Each Capacitor Since all the capacitors are in parallel, the voltage across each capacitor is the same: \[ V = 100 \, \text{V} \] The charge \( Q \) on each capacitor is given by: \[ Q = C \cdot V \] Thus, the charge on each capacitor is: \[ Q_1 = C_1 \cdot V = 2 \, \text{pF} \times 100 \, \text{V} = 200 \, \text{pC} \] \[ Q_2 = C_2 \cdot V = 3 \, \text{pF} \times 100 \, \text{V} = 300 \, \text{pC} \] \[ Q_3 = C_3 \cdot V = 4 \, \text{pF} \times 100 \, \text{V} = 400 \, \text{pC}

Question8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Solution :

Question9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Solution :(a) When the voltage supply remains connected:

The capacitance of the capacitor will become K times.

Therefore, C’ = kC

Where k = dielectric constant = 6×17.7pF = 106.2 pF

The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V

The charge on the capacitor,

q’ = C’V = 160.2 x 10-12 x 100

= 1.602 x 10-8 C

(b) After the voltage supply is disconnected:

As calculated above, the capacitance of the capacitor, C’ = 106.2 pF

The potential difference will decrease on introducing mica sheet by a factor of K,

Question10. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Solution :Given, C = 12 pF = 12 x 10-12 F

V = 50 V

The electrostatic energy stored in the capacitor,

W = (½) CV2 = (½) × 12 × 10-12× (50)2 = 1.5 × 10-8 J

Question11. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution :Given, C1 = 600 pF = 600 x 10-12 F

V1 = 200 V

Energy stored in the capacitor,

U1 = (½) C1 (V1)2 = (½)×600× 10-12× (200)2

= 12×10-6 J

When this charged capacitor is connected to another uncharged capacitor C2 ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.

Total charge on the two capacitors,

q = C1V1 + C2V2 = 600 × 10-12× 200 + 0

= 12 ×10-8 C

Total capacitance of the two capacitors,

C = C1 + C2 = 600 pF + 600 pF

= 1200 pF

= 1200 x 10-12 F

ADDITIONAL EXERCISES:

Question12. A charge of \(8 \, \text{mC}\) is located at the origin. Calculate the work done in taking a small charge of \(-2 \times 10^{-9} \, \text{C}\) from a point \(P (0, 0, 3 \, \text{cm})\) to a point \(Q (0, 4 \, \text{cm}, 0)\), via a point \(R (0, 6 \, \text{cm}, 9 \, \text{cm})\). \section

Solution- \textbf{Step 1:} Draw a rough diagram for the given situation. Let the origin be at point \(O\). \textbf{Step 2:} Find the potential at \(P\) and \(Q\). Given, Charge at the origin, \( q_1 = 8 \, \text{mC} = 8 \times 10^{-3} \, \text{C} \). Point \(P\) is at a distance \(O P = 3 \, \text{cm} = 0.03 \, \text{m}\) from the origin along the \(z\)-axis. Potential at \(P\) due to \(q_1\): \[ V_P = \frac{k q_1}{0.03} \] Point \(Q\) is at a distance \(O Q = 4 \, \text{cm} = 0.04 \, \text{m}\) from the origin along the \(y\)-axis. Potential at point \(Q\) due to \(q_1\): \[ V_Q = \frac{k q_1}{0.04} \] \textbf{Step 3:} Calculate the work done by the electrostatic force. The small charge being taken from point \(P\) to point \(Q\), with charge \(q = -2 \times 10^{-9} \, \text{C}\), and via point \(R\). Work done by the electrostatic force \(W\) is given by: \[ W = q (V_Q – V_P) \] Substituting the values: \[ W = \left(-2 \times 10^{-9}\right) \left(\frac{k q_1}{0.04} – \frac{k q_1}{0.03}\right) \] \[ W = k q_1 \left(-2 \times 10^{-9}\right) \left(\frac{1}{0.04} – \frac{1}{0.03}\right) \] \[ W = 9 \times 10^9 \times \left(-2 \times 10^{-9}\right) \times \left(8 \times 10^{-3}\right) \times \left(\frac{1}{0.04} – \frac{1}{0.03}\right) \] \[ W = 144 \times 10^{-3} \times \left(0.01\right) \] \[ W = 1.2 \, \text{J} \]

3. A cube of side \( a \) has a charge \( q \) at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. \section*

Solution \textbf{Step 1: Draw a rough figure of the given situation.} The cube of side \( a \) can be shown as follows: [Insert rough figure of the cube with charges at vertices and the center point marked] \textbf{Step 2: Find the distance of each vertex from the centre of the cube.} Given, the length of the side of a cube is \( a \). The distance from the center of the cube to a vertex can be calculated using the space diagonal of the cube, which is given by: \[ \text{Space diagonal} = a\sqrt{3} \] The distance from the center to one of the vertices (which is half of the space diagonal) is: \[ r = \frac{a\sqrt{3}}{2} \] \textbf{Step 3: Find the electric potential at the centre of the cube.} The electric potential at the centre of the cube due to the eight charges at the vertices is: \[ V = \sum_{i=1}^{8} \frac{kq}{r} \] \[ V = 8 \cdot \frac{kq}{r} \] Substituting the value of \( r \): \[ V = 8 \cdot \frac{kq}{\frac{a\sqrt{3}}{2}} \] \[ V = \frac{16kq}{a\sqrt{3}} \] \textbf{Step 4: Find the electric field at the centre of the cube.} The net electric field at the centre is the vector sum of the electric fields due to the individual charges. However, because the charges are symmetrically distributed around the center, the electric fields due to each charge cancel out. Thus, the electric field at the center of the cube is: \[ E = 0 \]

Question14. Two tiny spheres carrying charges \( 1.5 \, \mu C \) and \( 2.5 \, \mu C \) are located \( 30 \, \text{cm} \) apart. Find the potential and electric field: \begin{itemize} \item[(i)] at the mid-point of the line joining the two charges. \item[(ii)] at a point \( 10 \, \text{cm} \) from this midpoint in a plane normal to the line and passing through the mid-point. \end{itemize} \section*

{Solution} \textbf{Part (i)} \textbf{Step 1: Draw a rough figure of the given situation.} Two charges are placed at points \( A \) and \( B \) respectively. \( O \) is the midpoint of the line joining the two charges. \[ q_1 = 1.5 \, \mu C = 1.5 \times 10^{-6} \, C \] \[ q_2 = 2.5 \, \mu C = 2.5 \times 10^{-6} \, C \] \textbf{Step 2: Find electric potential at point \( O \).} The electric potential at the midpoint \( O \) due to charges at \( A \) and \( B \) is: \[ V_O = k \left(\frac{q_1}{d/2} + \frac{q_2}{d/2}\right) \] where \( d = 0.30 \, \text{m} \) is the distance between the charges. \[ V_O = 9 \times 10^9 \left(\frac{1.5 \times 10^{-6}}{0.15} + \frac{2.5 \times 10^{-6}}{0.15}\right) \] \[ V_O = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.15} \] \[ V_O = 2.4 \times 10^5 \, \text{V} \] \textbf{Step 3: Find the electric field at point \( O \).} Let the electric field at point \( O \) be \( E_O \). Since the charges at \( A \) and \( B \) are of opposite nature: \[ E_O = k \left(\frac{q_2}{(d/2)^2} – \frac{q_1}{(d/2)^2}\right) \] \[ E_O = 9 \times 10^9 \left(\frac{2.5 \times 10^{-6}}{(0.15)^2} – \frac{1.5 \times 10^{-6}}{(0.15)^2}\right) \] \[ E_O = 9 \times 10^9 \times \frac{1 \times 10^{-6}}{0.0225} \] \[ E_O = 4 \times 10^5 \, \text{V/m} \] The electric field is directed away from the larger charge. \textbf{Part (ii)} \textbf{Step 1: Draw a rough figure of the given situation.} Let a point \( Z \) such that the normal distance \( OZ = 10 \, \text{cm} = 0.1 \, \text{m} \). From the figure, the distance \( AZ = BZ = \sqrt{(AO)^2 + (OZ)^2} \) where \( AO = 0.15 \, \text{m} \) and \( OZ = 0.1 \, \text{m} \): \[ AZ = \sqrt{(0.1)^2 + (0.15)^2} = 0.18 \, \text{m} \] \textbf{Step 2: Find the electric potential at \( Z \).} The electric potential at point \( Z \) due to the charges at \( A \) and \( B \) is: \[ V_Z = k \left(\frac{q_1}{AZ} + \frac{q_2}{BZ}\right) \] \[ V_Z = 9 \times 10^9 \left(\frac{1.5 \times 10^{-6}}{0.18} + \frac{2.5 \times 10^{-6}}{0.18}\right) \] \[ V_Z = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.18} \] \[ V_Z = 2 \times 10^5 \, \text{V} \] \textbf{Step 3: Find the electric field at point \( Z \).} The electric field due to \( q_1 \) at \( Z \) is: \[ E_A = k \frac{q_1}{(AZ)^2} \] \[ E_A = 9 \times 10^9 \times \frac{1.5 \times 10^{-6}}{(0.18)^2} \] \[ E_A = 0.416 \times 10^6 \, \text{V/m} \] The electric field due to \( q_2 \) at \( Z \) is: \[ E_B = k \frac{q_2}{(BZ)^2} \] \[ E_B = 9 \times 10^9 \times \frac{2.5 \times 10^{-6}}{(0.18)^2} \] \[ E_B = 0.69 \times 10^6 \, \text{V/m} \] The resultant electric field intensity at \( Z \) is: \[ E = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos(2\theta)} \] From the figure, we can obtain: \[ \cos\theta = \frac{0.10}{0.18} = 0.555 \] \[ \cos(2\theta) = 2\cos^2\theta – 1 \] \[ \cos(2\theta) = 2 \times (0.555)^2 – 1 = -0.38 \] Hence, \[ E = \sqrt{(0.416 \times 10^6)^2 + (0.69 \times 10^6)^2 – 2 \times 0.416 \times 10^6 \times 0.69 \times 10^6 \times (-0.38)} \] \[ E = 6.6 \times 10^5 \, \text{V/m} \]

Question15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Solution :

Question16.

(b) Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Solution :(A)Electric field on one side of the charged body is E1 and the electric field on the other side of the same body be E2. If infinite plane charged body has uniform thickness then the electric field due to one surface of the body is given by,

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of the electrostatic field is continuous from one side of a charged surface to the other.

Question17. A long charged cylinder of linear charge density λ is surrounded by a hollow coaxial conducting cylinder. What is the electric field in the space between the two cylinders?

Solution :Given,

Charge density of the long-charged cylinder of length L and radius r is λ. Another same type of cylinder with radius R surrounded it.

Let E is the electric field produced in the space between the two cylinders.

Electric flux through a Gaussian surface is given by the Gaussian theorem as,

Φ = E(2πd)L

Where, d = distance of a point from the common axis of the cylinders.

Let q be the total charge on the cylinder,

Question18. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from the proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Solution :Given,

The distance between electron-proton of hydrogen atom = 0.53 Å = 0.53× 10-10m

Charge on electron, q1 = – 1.6×10-19 C

Charge on proton, q2 = 1.6×10-19 C

(a)Potential energy at infinity is Zero.

Potential energy of a system,

Question19.

Solution :

Question20. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Solution :Let a be the radius of the sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of the sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.

Question21. Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Solution :

Question21. Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Solution :

B. Given r/a>>1, which implies r>>a

the distance of the point where potential is to be obtained is much greater than half of the distance between the two charges.

Hence, the potential (V) at a distance r is inversely proportional to square of the distance, i.e. V ∞1/r2

Question22. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on \( r \) for \( r/a \gg 1 \), and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge). \section

Solution- Four charges of the same magnitude are placed at points X, Y, Y, and Z, respectively, as shown in the following figure. A point is located at P, which is \( r \) distance away from point Y. The system of charges forms an electric quadrupole. It can be considered that the system of the electric quadrupole has three charges: – Charge \( +q \) placed at point X – Charge \( -2q \) placed at point Y – Charge \( +q \) placed at point Z Given: \[ \text{XY} = \text{YZ} = a \] \[ \text{YP} = r \] \[ \text{PX} = r + a \] \[ \text{PZ} = r – a \] The electrostatic potential \( V \) caused by the system of three charges at point P is given by: \[ V = k \left(\frac{q}{r+a} – \frac{2q}{r} + \frac{q}{r-a}\right) \] For \( r \gg a \), we can expand the terms in the series and keep the leading terms: \[ V \approx kq \left(\frac{1}{r+a} – \frac{2}{r} + \frac{1}{r-a}\right) \] Using binomial expansion: \[ \frac{1}{r\pm a} \approx \frac{1}{r} \mp \frac{a}{r^2} \] Substituting these into the expression for \( V \): \[ V \approx kq \left(\frac{1}{r} – \frac{a}{r^2} – \frac{2}{r} + \frac{1}{r} + \frac{a}{r^2}\right) \] \[ V \approx -\frac{kqa^2}{r^3} \] Hence, the potential due to an electric quadrupole is inversely proportional to the cube of the distance from the quadrupole, i.e., \( V \propto \frac{1}{r^3} \). **Comparison:** – For an electric dipole, the potential varies as \( V \propto \frac{1}{r^2} \). – For an electric monopole (a single charge), the potential varies as \( V \propto \frac{1}{r} \). This indicates that the potential falls off more rapidly with distance for a quadrupole than for a dipole or monopole.

Question23. An electrical technician requires a capacitance of \( 2 \mu F \) in a circuit across a potential difference of \( 1 \, \text{kV} \). A large number of \( 1 \mu F \) capacitors are available to him, each of which can withstand a potential difference of not more than \( 400 \, \text{V} \). Suggest a possible arrangement that requires the minimum number of capacitors. \section*

{Solution- \textbf{Step 1: Find the minimum number of capacitors in series} Given: \begin{align*} \text{Required Capacitance, } & C = 2 \, \mu F \\ \text{Potential difference, } & V = 1 \, \text{kV} = 1000 \, \text{V} \\ \text{Capacitance of each capacitor, } & C’ = 1 \, \mu F \\ \text{Potential difference that each capacitor can withstand, } & V’ = 400 \, \text{V} \end{align*} Suppose \( m \) capacitors are connected in series. The potential difference across each capacitor is given by: \[ \frac{V}{m} = V’ \] \[ m = \frac{1000}{400} = 2.5 \approx 3 \] Therefore, the number of capacitors connected in series is three (in one row). \textbf{Step 2: Find the effective capacitance of one row} The effective capacitance of one row is: \[ \frac{1}{C’} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \] \[ C’ = \frac{1}{3} \, \mu F \quad \dots (i) \] \textbf{Step 3: Find the number of rows required in parallel} Let there be \( n \) parallel rows, and we know each of these rows will have 3 capacitors. Therefore, the equivalent capacitance of the circuit is given as: \[ C” = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \ldots n \text{ times} \] \[ C” = \frac{n}{3} \quad \dots (ii) \] Since the required capacitance of the circuit is \( 2 \, \mu F \), from equation (ii) we get: \[ 2 = \frac{n}{3} \] \[ n = 2 \times 3 = 6 \] Hence, there are 6 rows of three capacitors in the circuit. \textbf{Step 4: Find the number of capacitors for the required arrangement} The minimum number of capacitors required for the arrangement is: \[ N = m \times n = 3 \times 6 \] \[ N = 18 \] \textbf{Final Answer:} 18 capacitors are required for the possible arrangement.

Question24. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Solution:

Question25. \documentclass{article} \usepackage{amsmath} \begin{document} \section*{Question} Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor. \section*

{Solution} Given: The supply voltage is 300 V. Consider the given network shown below. \begin{figure}[h!] \centering % Include the figure of the network here \caption{Capacitor network} \end{figure} \textbf{Step 1: Calculate the equivalent capacitance of capacitors $C_2$ and $C_3$} Since the capacitors $C_2$ and $C_3$ are connected in series, the equivalent capacitance $C’$ of these two capacitors is given by: \[ \frac{1}{C’} = \frac{1}{C_2} + \frac{1}{C_3} \] Substituting the given values: \[ \frac{1}{C’} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} \] \[ C’ = 100\, \text{pF} \] \textbf{Step 2: Calculate the equivalent capacitance of $C_1$ and $C’$} Since the capacitors $C_1$ and $C’$ are connected in parallel, the equivalent capacitance $C”$ of these two capacitors is given by: \[ C” = C’ + C_1 \] Substituting the given values: \[ C” = 100 + 100 = 200\, \text{pF} \] \textbf{Step 3: Calculate the equivalent capacitance of $C”$ and $C_4$} Since the capacitors $C”$ and $C_4$ are connected in series, the equivalent capacitance $C$ of these two capacitors is given by: \[ \frac{1}{C} = \frac{1}{C”} + \frac{1}{C_4} \] Substituting the given values: \[ \frac{1}{C} = \frac{1}{200} + \frac{1}{100} = \frac{3}{200} \] \[ C = \frac{200}{3}\, \text{pF} \] \textbf{Step 4: Calculate the charge and voltage across each capacitor} The total voltage is 300 V, distributed between $C”$ and $C_4$. The charge on $C_4$ is given by: \[ Q_4 = C \times V \] \[ Q_4 = \frac{200}{3} \times 10^{-12} \times 300 = 2 \times 10^{-8}\, \text{C} \] The potential difference across $C_4$ is: \[ V_4 = \frac{Q_4}{C_4} \] \[ V_4 = \frac{2 \times 10^{-8}}{100 \times 10^{-12}} = 200\, \text{V} \] The voltage across $C_1$ is: \[ V_1 = V – V_4 \] \[ V_1 = 300 – 200 = 100\, \text{V} \] The charge on $C_1$ is: \[ Q_1 = C_1 \times V_1 \] \[ Q_1 = 100 \times 10^{-12} \times 100 = 1 \times 10^{-8}\, \text{C} \] Since $C_2$ and $C_3$ are in series, the voltage across each is: \[ V_2 = V_3 = \frac{V_1}{2} = 50\, \text{V} \] The charge on $C_2$ and $C_3$ is: \[ Q_2 = C_2 \times V_2 = 1 \times 10^{-8}\, \text{C} \] \[ Q_3 = C_3 \times V_3 = 1 \times 10^{-8}\, \text{C} \] Thus, the equivalent capacitance of the given circuit is $\frac{200}{3}\, \text{pF}$, and the charges and voltages across each capacitor are: \[ \begin{aligned} &Q_1 = 10^{-8}\, \text{C}, \quad V_1 = 100\, \text{V} \\ &Q_2 = 10^{-8}\, \text{C}, \quad V_2 = 50\, \text{V} \\ &Q_3 = 10^{-8}\, \text{C}, \quad V_3 = 50\, \text{V} \\ &Q_4 = 2 \times 10^{-8}\, \text{C}, \quad V_4 = 200\, \text{V} \end{aligned} \]

Question26. The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Solution :

Question27. A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Solution :

∴ U = 5.33×10-2 J

Thus, the electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation = 8×10-2 J-5.33×10-2 J = 6.67×10-2J

Question28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.

Solution :

Question29.

Solution:

Question30. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Solution :

Radius of the outer shell (r1) = 13cm = 0.13m

Radius of the inner shell (r2) = 12cm = 0.12m

Charge on the outer surface of the inner shell = 2.5 μC = 2.5×10-6C

Dielectric constant of liquid = 32

Since, Potential difference between the two shells,

Question31. Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1 Q2/4π∈or 2, where r is the distance between their centres?

(b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Solution :
(a)

(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of1/r2, on r.

(c) Yes,

If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

(e) No

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

Question32. A cylindrical capacitor has two coaxial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Solution :Radius of outer cylinder(R) = 1.5cm = 0.015m

Radius of inner cylinder(r) = 1.4cm = 0.014m

Charge on the inner cylinder(q) = 3.5µC = 3.5×10-6C

Question33. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Solution :Potential difference of a parallel plate capacitor(V) = 1kV = 1000V

Dielectric constant of a material(ϵr) = 3

Dielectric strength = 107V/m

Electric field intensity(E) = 10%of 107

⇒ E = 106V/m

(since, the field intensity never exceeds 10% of the dielectric strength)

Question34. Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Solution :
(a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.

(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) Concentric spheres centered at the origin are equipotential surfaces.

(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Question35. In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Solution :
Potential difference, V = 15 × 106 V

Dielectric strength of the surrounding gas = 5 × 107 V/m

Electric field intensity, E = Dielectric strength = 5 × 107 V/m

Minimum radius of the spherical shell required for the purpose is given by,

Hence, the minimum radius of the spherical shell required is 30 cm.

Question36. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

Solution :
According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.

Question37. Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning strike? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is a good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Solution :
(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of a discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

(d) During lightning and thunderstorms, light energy, heat energy, and sound energy are dissipated in the atmosphere.