CBSE Class 12 Physics Chapter 3: Electrostatic Potential and Capacitance - NCERT Solutions

The work done in moving a charge in an electrostatic field is independent of the path taken. It only depends on the initial and final positions and the charges involved.

Given:

• Charge at the origin, $q = 8 \text{ mC} = 8 \times 10^{-3} \text{ C}$.
• Small charge to be moved, $q_1 = -2 \times 10^{-9} \text{ C}$.
• Initial point P is at (0, 0, 3 cm), so its distance from the origin is $d_1 = 3 \text{ cm} = 0.03 \text{ m}$.
• Final point Q is at (0, 4 cm, 0), so its distance from the origin is $d_2 = 4 \text{ cm} = 0.04 \text{ m}$.

The electrostatic potential at a distance $d$ from a charge $q$ is given by $V = \frac{q}{4\pi \epsilon_0 d}$.

The potential at point P is:

$V_P = \frac{q}{4\pi \epsilon_0 d_1}$

The potential at point Q is:

$V_Q = \frac{q}{4\pi \epsilon_0 d_2}$

The work done ($W$) in moving the charge $q_1$ from P to Q is given by:

$W = q_1 (V_Q - V_P)$

$W = q_1 \left( \frac{q}{4\pi \epsilon_0 d_2} - \frac{q}{4\pi \epsilon_0 d_1} \right)$

$W = \frac{qq_1}{4\pi \epsilon_0} \left( \frac{1}{d_2} - \frac{1}{d_1} \right)$

We know that $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$.

Substituting the values:

$W = (9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}) \times (8 \times 10^{-3} \text{ C}) \times (-2 \times 10^{-9} \text{ C}) \left( \frac{1}{0.04 \text{ m}} - \frac{1}{0.03 \text{ m}} \right)$

$W = (72 \times 10^{-6}) \times (-2 \times 10^{-9}) \left( 25 - \frac{100}{3} \right) \text{ J}$

$W = -144 \times 10^{-15} \left( \frac{75 - 100}{3} \right) \text{ J}$

$W = -144 \times 10^{-15} \left( \frac{-25}{3} \right) \text{ J}$

$W = 48 \times 25 \times 10^{-15} \text{ J}$

$W = 1200 \times 10^{-15} \text{ J} = 1.2 \times 10^{-12} \text{ J}$

Let's recheck the calculation with the provided source values:

$W = 9 \times 10^9 \times 8 \times 10^{-3} \times (-2 \times 10^{-9}) \left[\frac{1}{0.04} - \frac{1}{0.03}\right]$

$W = -144 \times 10^{-3} \left[ 25 - \frac{100}{3} \right]$

$W = -144 \times 10^{-3} \left[ \frac{75 - 100}{3} \right]$

$W = -144 \times 10^{-3} \left[ \frac{-25}{3} \right]$

$W = 48 \times 25 \times 10^{-3} \text{ J}$

$W = 1200 \times 10^{-3} \text{ J} = 1.2 \text{ J}$

There seems to be a discrepancy in the source's final calculation. Let's use the intermediate step provided in the source to verify:

$W = -144\times10^{-3}\times\left(\frac{-25}{3}\right) = 48 \times 25 \times 10^{-3} = 1200 \times 10^{-3} = 1.2 \text{ J}$

The source states 1.27 J, which might be due to rounding or a slight error in the source's calculation. Based on the provided intermediate steps, the result is 1.2 J.

Answer: The work done during the process is 1.2 J.

Consider a cube of side length $b$. There are 8 vertices, and each vertex has a charge $q$. We need to find the electric potential and electric field at the center of the cube.

First, let's find the distance from each vertex to the center of the cube. The length of the space diagonal of the cube is $l = b\sqrt{3}$. The distance from the center to any vertex is half of the space diagonal.

Distance $r = \frac{l}{2} = \frac{b\sqrt{3}}{2}$.

Electric Potential at the center:

The electric potential at the center is the algebraic sum of the potentials due to all eight charges. Since the distance from each vertex to the center is the same ($r$), and each vertex has the same charge ($q$), the potential due to each charge is $V_{vertex} = \frac{1}{4\pi \epsilon_0} \frac{q}{r}$.

The total potential $V$ at the center is:

$V = 8 \times V_{vertex} = 8 \times \frac{1}{4\pi \epsilon_0} \frac{q}{r}$

$V = 8 \times \frac{1}{4\pi \epsilon_0} \frac{q}{\frac{b\sqrt{3}}{2}}$

$V = \frac{8 \times 2}{4\pi \epsilon_0} \frac{q}{b\sqrt{3}} = \frac{16}{4\pi \epsilon_0} \frac{q}{b\sqrt{3}}$

$V = \frac{4}{\pi \epsilon_0} \frac{q}{b\sqrt{3}} = \frac{4q}{\sqrt{3}\pi \epsilon_0 b}$

Electric Field at the center:

The electric field at the center is the vector sum of the electric fields due to all eight charges. Due to the symmetry of the cube, the electric field contributions from charges at opposite vertices cancel each other out. For every charge, there is an equal and opposite charge at the diametrically opposite vertex, and the distance to the center is the same. Therefore, the net electric field at the center of the cube is zero.

Answer: The potential at the centre of the cube is $\frac{4q}{\sqrt{3}\pi \epsilon_0 b}$ and the electric field at the centre is zero.

Given:

• Charge $q_1 = 1.5 \text{ µC} = 1.5 \times 10^{-6} \text{ C}$.
• Charge $q_2 = 2.5 \text{ µC} = 2.5 \times 10^{-6} \text{ C}$.
• Distance between the charges, $d = 30 \text{ cm} = 0.30 \text{ m}$.
• The constant $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$.

Part (a): At the mid-point of the line joining the two charges.

Let the mid-point be O. The distance of the mid-point from each charge is $\frac{d}{2}$.

Distance from $q_1$ to O = $\frac{0.30}{2} = 0.15 \text{ m}$.

Distance from $q_2$ to O = $\frac{0.30}{2} = 0.15 \text{ m}$.

Potential at the mid-point ($V_O$):

The potential at O is the algebraic sum of the potentials due to $q_1$ and $q_2$.

$V_O = \frac{1}{4\pi \epsilon_0} \frac{q_1}{(\frac{d}{2})} + \frac{1}{4\pi \epsilon_0} \frac{q_2}{(\frac{d}{2})}$

$V_O = \frac{1}{4\pi \epsilon_0 \frac{d}{2}} (q_1 + q_2)$

$V_O = \frac{2}{4\pi \epsilon_0 d} (q_1 + q_2)$

$V_O = \frac{2 \times (9 \times 10^9 \text{ N m}^2 \text{ C}^{-2})}{(0.30 \text{ m})} (1.5 \times 10^{-6} \text{ C} + 2.5 \times 10^{-6} \text{ C})$

$V_O = \frac{18 \times 10^9}{0.30} (4.0 \times 10^{-6}) \text{ V}$

$V_O = 60 \times 10^9 \times 4.0 \times 10^{-6} \text{ V}$

$V_O = 240 \times 10^3 \text{ V} = 2.4 \times 10^5 \text{ V}$

Electric Field at the mid-point ($E_O$):

The electric field at O is the vector sum of the fields due to $q_1$ and $q_2$. The field due to $q_1$ is directed away from $q_1$ (towards $q_2$), and the field due to $q_2$ is directed away from $q_2$ (towards $q_1$). Since $q_2 > q_1$, the net field will be directed towards $q_2$.

Magnitude of electric field due to $q_1$ at O:

$E_1 = \frac{1}{4\pi \epsilon_0} \frac{q_1}{(\frac{d}{2})^2} = \frac{9 \times 10^9 \times 1.5 \times 10^{-6}}{(0.15)^2} = \frac{13.5 \times 10^3}{0.0225} = 600 \times 10^3 \text{ N/C}$

Magnitude of electric field due to $q_2$ at O:

$E_2 = \frac{1}{4\pi \epsilon_0} \frac{q_2}{(\frac{d}{2})^2} = \frac{9 \times 10^9 \times 2.5 \times 10^{-6}}{(0.15)^2} = \frac{22.5 \times 10^3}{0.0225} = 1000 \times 10^3 \text{ N/C}$

Since both fields are along the line joining the charges and point in opposite directions, the net electric field $E_O$ is the difference between their magnitudes. The direction is towards the larger charge ($q_2$).

$E_O = E_2 - E_1 = (1000 - 600) \times 10^3 \text{ N/C} = 400 \times 10^3 \text{ N/C} = 4.0 \times 10^5 \text{ N/C}$

Part (b): At a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Let the mid-point be O. Let the point be P, which is 10 cm (0.1 m) away from O in a plane normal to the line joining the charges. The distance $r = 0.1 \text{ m}$.

The distance from $q_1$ to P is $r_1 = \sqrt{(\frac{d}{2})^2 + r^2} = \sqrt{(0.15)^2 + (0.1)^2} = \sqrt{0.0225 + 0.01} = \sqrt{0.0325} \text{ m}$.

The distance from $q_2$ to P is $r_2 = \sqrt{(\frac{d}{2})^2 + r^2} = \sqrt{(0.15)^2 + (0.1)^2} = \sqrt{0.0325} \text{ m}$.

So, $r_1 = r_2 = \sqrt{0.0325} \text{ m}$.

Potential at point P ($V_P$):

The potential at P is the algebraic sum of potentials due to $q_1$ and $q_2$.

$V_P = \frac{1}{4\pi \epsilon_0} \frac{q_1}{r_1} + \frac{1}{4\pi \epsilon_0} \frac{q_2}{r_2}$

$V_P = \frac{1}{4\pi \epsilon_0} \frac{1}{\sqrt{0.0325}} (q_1 + q_2)$

$V_P = \frac{9 \times 10^9}{\sqrt{0.0325}} (1.5 \times 10^{-6} + 2.5 \times 10^{-6}) \text{ V}$

$V_P = \frac{9 \times 10^9}{\sqrt{0.0325}} (4.0 \times 10^{-6}) \text{ V}$

$V_P = \frac{36 \times 10^3}{\sqrt{0.0325}} \text{ V} \approx \frac{36000}{0.1803} \text{ V} \approx 199667 \text{ V} \approx 2.0 \times 10^5 \text{ V}$

Electric Field at point P ($E_P$):

The electric field at P is the vector sum of the fields due to $q_1$ and $q_2$. Let $E_1$ be the field due to $q_1$ and $E_2$ be the field due to $q_2$. The magnitudes are:

$E_1 = \frac{1}{4\pi \epsilon_0} \frac{q_1}{r_1^2} = \frac{9 \times 10^9 \times 1.5 \times 10^{-6}}{0.0325} = \frac{13.5 \times 10^3}{0.0325} \approx 415385 \text{ N/C}$

$E_2 = \frac{1}{4\pi \epsilon_0} \frac{q_2}{r_2^2} = \frac{9 \times 10^9 \times 2.5 \times 10^{-6}}{0.0325} = \frac{22.5 \times 10^3}{0.0325} \approx 692308 \text{ N/C}$

The electric field components along the line joining the charges will cancel out due to symmetry. The components perpendicular to the line joining the charges will add up. Let $\theta$ be the angle between the line joining $q_1$ to P and the line joining O to P. Then $\cos\theta = \frac{0.15}{\sqrt{0.0325}}$ and $\sin\theta = \frac{0.1}{\sqrt{0.0325}}$.

The net electric field $E_P$ will be in the direction perpendicular to the line joining the charges (along the direction OP).

$E_P = E_1 \sin\theta + E_2 \sin\theta = (E_1 + E_2) \sin\theta$

$E_P = \left( \frac{1}{4\pi \epsilon_0} \frac{q_1}{r_1^2} + \frac{1}{4\pi \epsilon_0} \frac{q_2}{r_2^2} \right) \sin\theta$

Since $r_1 = r_2$, let $r = \sqrt{0.0325}$.

$E_P = \frac{1}{4\pi \epsilon_0} \frac{(q_1 + q_2)}{r^2} \sin\theta$

$E_P = \frac{9 \times 10^9 \times (4.0 \times 10^{-6})}{0.0325} \times \frac{0.1}{\sqrt{0.0325}}$

$E_P = \frac{36 \times 10^3}{0.0325} \times \frac{0.1}{\sqrt{0.0325}} \approx 1.107 \times 10^6 \times 0.555 \approx 6.14 \times 10^5 \text{ N/C}$

Answer:

(a) At the mid-point: Potential = $2.4 \times 10^5 \text{ V}$, Electric Field = $4.0 \times 10^5 \text{ N/C}$ (directed towards the charge 2.5 µC).

(b) At a point 10 cm from the midpoint in the perpendicular plane: Potential $\approx 2.0 \times 10^5 \text{ V}$, Electric Field $\approx 6.14 \times 10^5 \text{ N/C}$ (directed away from the midpoint, perpendicular to the line joining the charges).