CBSE Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance - NCERT Additional Exercises Solutions

We are given a source charge $q = 8 \text{ mC} = 8 \times 10^{-3} \text{ C}$ located at the origin. We need to find the work done in moving a test charge $q_1 = -2 \times 10^{-9} \text{ C}$ from point P to point Q.

Point P is at a distance $d_1 = 3 \text{ cm} = 0.03 \text{ m}$ from the origin along the z-axis. Its coordinates are (0, 0, 0.03).

Point Q is at a distance $d_2 = 4 \text{ cm} = 0.04 \text{ m}$ from the origin along the y-axis. Its coordinates are (0, 0.04, 0).

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

The potential at point P due to the charge at the origin is:

$$V\_P\; =\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; d\_1\}$$

The potential at point Q due to the charge at the origin is:

$$V\_Q\; =\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; d\_2\}$$

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

$$W\; =\; q\_1\; (V\_Q\; -\; V\_P)$$

Substituting the expressions for $V_Q$ and $V_P$:

$$W\; =\; q\_1\; \backslash left(\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; d\_2\}\; -\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; d\_1\}\; \backslash right)\; =\; \backslash frac\{qq\_1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash left(\; \backslash frac\{1\}\{d\_2\}\; -\; \backslash frac\{1\}\{d\_1\}\; \backslash right)$$

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

Now, substitute the given values:

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

$$W\; =\; (9\; \backslash times\; 8\; \backslash times\; -2)\; \backslash times\; 10^\{(9\; -\; 3\; -\; 9)\}\; \backslash left(\; 25\; -\; \backslash frac\{100\}\{3\}\; \backslash right)\; \backslash text\{\; J\}$$

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

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

$$W\; =\; \backslash frac\{144\; \backslash times\; 25\}\{3\}\; \backslash times\; 10^\{-3\}\; \backslash text\{\; J\}$$

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

$$W\; =\; 1200\; \backslash times\; 10^\{-3\}\; \backslash text\{\; J\}\; =\; 1.20\; \backslash text\{\; J\}$$

Note: The path via point R is irrelevant as electrostatic work done is path-independent.

Therefore, the work done during the process is 1.20 J.

Consider a cube of side length $b$. There are 8 vertices, and each vertex has a charge $q$ placed on it.

We need to find the electric potential (V) and the electric field (E) at the center of the cube.

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

So, the distance $r$ from the center to each vertex is:

$$r\; =\; \backslash frac\{l\}\{2\}\; =\; \backslash frac\{b\backslash sqrt\{3\}\}\{2\}$$

Electric Potential (V) at the center:

The electric potential at the center is the algebraic sum of the potentials due to all 8 charges located at the vertices. Since the distance from the center to each vertex is the same ($r$), and each charge is $q$, the potential due to one charge is $V_{single} = \frac{q}{4\pi \epsilon_0 r}$.

The total potential $V$ at the center is the sum of potentials from all 8 charges:

$$V\; =\; 8\; \backslash times\; V\_\{single\}\; =\; 8\; \backslash times\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; r\}$$

Substitute $r = \frac{b\sqrt{3}}{2}$:

$$V\; =\; 8\; \backslash times\; \backslash frac\{q\}\{4\backslash pi\; \backslash epsilon\_0\; \backslash left(\backslash frac\{b\backslash sqrt\{3\}\}\{2\}\backslash right)\}\; =\; \backslash frac\{8q\; \backslash times\; 2\}\{4\backslash pi\; \backslash epsilon\_0\; b\backslash sqrt\{3\}\}\; =\; \backslash frac\{16q\}\{4\backslash pi\; \backslash epsilon\_0\; b\backslash sqrt\{3\}\}\; =\; \backslash frac\{4q\}\{\backslash pi\; \backslash epsilon\_0\; b\backslash sqrt\{3\}\}$$

Therefore, the potential at the center of the cube is $\frac{4q}{\pi \epsilon_0 b\sqrt{3}}$.

Electric Field (E) at the center:

The electric field at the center is the vector sum of the electric fields due to all 8 charges at the vertices. Due to the symmetry of the cube, for every charge $q$ at a vertex, there is an identical charge $q$ at the diagonally opposite vertex. The electric field produced by these two charges at the center will be equal in magnitude but opposite in direction. Therefore, they cancel each other out.

Since this applies to all four pairs of diagonally opposite vertices, the net electric field at the center of the cube is zero.

$$E\_\{center\}\; =\; 0$$

Given charges are $q_1 = 1.5 \text{ µC} = 1.5 \times 10^{-6} \text{ C}$ and $q_2 = 2.5 \text{ µC} = 2.5 \times 10^{-6} \text{ C}$.

The distance between the charges is $d = 30 \text{ cm} = 0.30 \text{ m}$.

The electrostatic constant is $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$.

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

Let O be the mid-point. The distance of O from each charge is $\frac{d}{2} = \frac{0.30}{2} = 0.15 \text{ m}$.

Potential (V) at the mid-point:

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

$$V\; =\; V\_1\; +\; V\_2\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_1\}\{(d/2)\}\; +\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_2\}\{(d/2)\}$$

$$V\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\; \backslash left(d/2\backslash right)\}\; (q\_1\; +\; q\_2)$$

$$V\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (1.5\; \backslash times\; 10^\{-6\}\; +\; 2.5\; \backslash times\; 10^\{-6\})\}\{0.15\}$$

$$V\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (4.0\; \backslash times\; 10^\{-6\})\}\{0.15\}\; =\; \backslash frac\{36\; \backslash times\; 10^3\}\{0.15\}\; =\; \backslash frac\{36000\}\{0.15\}\; =\; 240000\; \backslash text\{\; V\}$$

$$V\; =\; 2.4\; \backslash times\; 10^5\; \backslash text\{\; V\}$$

Electric Field (E) at the mid-point:

The electric field due to $q_1$ at O is $E_1 = \frac{1}{4\pi \epsilon_0} \frac{q_1}{(d/2)^2}$ (directed away from $q_1$).

The electric field due to $q_2$ at O is $E_2 = \frac{1}{4\pi \epsilon_0} \frac{q_2}{(d/2)^2}$ (directed away from $q_2$).

Since $q_2 > q_1$, $E_2 > E_1$. Both fields are directed towards the right (from $q_1$ to $q_2$).

The net electric field $E$ at O is the difference between $E_2$ and $E_1$ (since they are in the same direction):

$$E\; =\; E\_2\; -\; E\_1\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_2\}\{(d/2)^2\}\; -\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_1\}\{(d/2)^2\}$$

$$E\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\; \backslash left(d/2\backslash right)^2\}\; (q\_2\; -\; q\_1)$$

$$E\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (2.5\; \backslash times\; 10^\{-6\}\; -\; 1.5\; \backslash times\; 10^\{-6\})\}\{(0.15)^2\}$$

$$E\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (1.0\; \backslash times\; 10^\{-6\})\}\{0.0225\}\; =\; \backslash frac\{9\; \backslash times\; 10^3\}\{0.0225\}\; =\; \backslash frac\{9000\}\{0.0225\}\; =\; 400000\; \backslash text\{\; N/C\}$$

$$E\; =\; 4.0\; \backslash times\; 10^5\; \backslash text\{\; N/C\}$$

The direction of the net electric field is from charge $q_1$ towards charge $q_2$.

b) At a point 10 cm from the mid-point in a plane normal to the line joining the charges:

Let the mid-point be O. Let the point be P, such that OP = $x = 10 \text{ cm} = 0.10 \text{ m}$. The line joining the charges is along the x-axis, and P is on the y-axis (or any axis perpendicular to the line joining charges, passing through O).

The distance from $q_1$ to P is $r_1 = \sqrt{(d/2)^2 + x^2}$.

The distance from $q_2$ to P is $r_2 = \sqrt{(d/2)^2 + x^2}$.

So, $r_1 = r_2 = \sqrt{(0.15)^2 + (0.10)^2} = \sqrt{0.0225 + 0.01} = \sqrt{0.0325} \text{ m}$.

Potential (V) at point P:

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

$$V\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_1\}\{r\_1\}\; +\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{q\_2\}\{r\_2\}\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{(q\_1\; +\; q\_2)\}\{r\_1\}$$

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

$$V\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (4.0\; \backslash times\; 10^\{-6\})\}\{\backslash sqrt\{0.0325\}\}\; =\; \backslash frac\{36\; \backslash times\; 10^3\}\{\backslash sqrt\{0.0325\}\}\; \backslash approx\; \backslash frac\{36000\}\{0.1803\}\; \backslash approx\; 199667\; \backslash text\{\; V\}$$

$$V\; \backslash approx\; 2.0\; \backslash times\; 10^5\; \backslash text\{\; V\}$$

Electric Field (E) at point P:

The electric field $E_1$ due to $q_1$ and $E_2$ due to $q_2$ will have components along the line joining the charges and perpendicular to it. Due to symmetry, the components perpendicular to the line joining the charges will cancel out.

The component of $E_1$ along the line joining the charges is $E_{1x} = E_1 \cos \theta_1 = \frac{1}{4\pi \epsilon_0} \frac{q_1}{r_1^2} \frac{(d/2)}{r_1} = \frac{1}{4\pi \epsilon_0} \frac{q_1 (d/2)}{r_1^3}$.

The component of $E_2$ along the line joining the charges is $E_{2x} = E_2 \cos \theta_2 = \frac{1}{4\pi \epsilon_0} \frac{q_2}{r_2^2} \frac{(d/2)}{r_2} = \frac{1}{4\pi \epsilon_0} \frac{q_2 (d/2)}{r_2^3}$.

Since $r_1 = r_2$, the net field along the line joining the charges is:

$$E\_x\; =\; E\_\{1x\}\; +\; E\_\{2x\}\; =\; \backslash frac\{1\}\{4\backslash pi\; \backslash epsilon\_0\}\; \backslash frac\{(q\_1\; +\; q\_2)(d/2)\}\{r\_1^3\}$$

$$E\_x\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (4.0\; \backslash times\; 10^\{-6\})\; \backslash times\; (0.15)\}\{(\backslash sqrt\{0.0325\})^3\}\; =\; \backslash frac\{54000\}\{(0.0325)^\{3/2\}\}\; \backslash approx\; \backslash frac\{54000\}\{0.0584\}\; \backslash approx\; 924657\; \backslash text\{\; N/C\}$$

$$E\_x\; \backslash approx\; 9.25\; \backslash times\; 10^5\; \backslash text\{\; N/C\}$$

The direction of the net electric field is along the line joining the charges, pointing from $q_1$ towards $q_2$.