## Two point charges 4q and -q are fixed on the x-axis at x=-d/2 and x= +d/2 respectively ….

Q: Two point charges 4q and -q are fixed on the x-axis at x=-d/2 and x= +d/2 respectively . If a third point charge q is taken from origin to x = d along the semicircle as shown in the figure , the energy of charge will (a) increase by $\displaystyle \frac{3 q^2}{4\pi\epsilon_0 d}$

(b) decrease by $\displaystyle \frac{q^2}{4\pi\epsilon_0 d}$

(c) decrease by $\displaystyle \frac{4 q^2}{3\pi\epsilon_0 d}$

(d) increase by $\displaystyle \frac{2 q^2}{3\pi\epsilon_0 d}$

Click to See Solution :
Ans: (c)

Sol: Electrostatic Potential Energy when charge q is at origin O

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} [\frac{4q \times q}{d/2} + \frac{(-q) \times q}{d/2}]$

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} \frac{6q^2}{d}$

Electrostatic Potential Energy when charge q moved to x = d

$\displaystyle U_2 = \frac{1}{4\pi\epsilon_0} [\frac{4q \times q}{3d/2} + \frac{(-q) \times q}{d/2}]$

$\displaystyle U_1 = \frac{1}{4\pi\epsilon_0} \frac{2q^2}{3d}$

The change in Poential Energy ,

$\displaystyle \Delta U = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d} (\frac{2}{3}-6)$

$\displaystyle \Delta U = – \frac{4 q^2}{3\pi\epsilon_0 d}$

## A charged particle (mass m charge q) moves along X-axis with velocity vo . When it passes through origin it enters …

Q: A charged particle (mass m charge q) moves along X-axis with velocity vo . When it passes through origin it enters a region having uniform electric field $\vec{E} = -E\hat{j}$ which extends up to x = d . Equation of path of electron in the region x > d is (a) $\displaystyle y = \frac{q E d}{m v_o^2}(x-d)$

(b) $\displaystyle y = \frac{q E d}{m v_o^2}(\frac{d}{2}-x)$

(c) $\displaystyle y = \frac{q E d}{m v_o^2} x$

(d) $\displaystyle y = \frac{q E d^2}{m v_o^2} x$

Click to See Solution :
Ans: (b)
Sol:

## Consider two charged metallic spheres S1 and S2 of radii R1 and R2 respectively . The electric field E1 (on S1) and E2 (on S2) …..

Q: Consider two charged metallic spheres S1 and S2 of radii R1 and R2 respectively . The electric field E1 (on S1) and E2 (on S2) on their surfaces are such that $\frac{E_1}{E_2} = \frac{R_1}{R_2}$ . Then the ratio V1 (on S1)/V2(on S2) of the electrostatic potentials on each sphere is

(a) $\displaystyle (\frac{R_1}{R_2})^3$

(b) $\displaystyle \frac{R_2}{R_1}$

(c) $\displaystyle \frac{R_1}{R_2}$

(d) $\displaystyle (\frac{R_1}{R_2})^2$

Click to See Solution :
Ans: (d)
Sol: $\displaystyle \frac{E_1}{E_2} = \frac{R_1}{R_2}$ (given) …(i)

$\displaystyle E_1 = K\frac{Q_1}{R_1^2}$

$\displaystyle E_2 = K\frac{Q_2}{R_2^2}$

$\displaystyle \frac{E_1}{E_2} = \frac{Q_1}{Q_2} \times \frac{R_2^2}{R_1^2}$

$\displaystyle \frac{R_1}{R_2} = \frac{Q_1}{Q_2} \times \frac{R_2^2}{R_1^2}$ (from (i))

$\displaystyle \frac{Q_1}{Q_2} = \frac{R_1^3}{R_2^3}$ …(ii)

$\displaystyle V_1 = K\frac{Q_1}{R_1}$

$\displaystyle V_2 = K\frac{Q_2}{R_2}$

$\displaystyle \frac{V_1}{V_2} = \frac{Q_1}{Q_2} \times \frac{R_2}{R_1}$

$\displaystyle \frac{V_1}{V_2} = \frac{R_1^3}{R_2^3} \times \frac{R_2}{R_1}$ (from (ii))

$\displaystyle \frac{V_1}{V_2} = \frac{R_1^2}{R_2^2}$

## A capacitor is made of two square plates each of side ‘ a ‘ making a very small angle α between them ….

Q: A capacitor is made of two square plates each of side ‘ a ‘ making a very small angle α between them , as shown in figure .The capacitance will be close to (a) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{4d})$

(b) $\displaystyle \frac{\epsilon_0 a^2}{d}(1 +\frac{\alpha a}{4d})$

(c) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{2d})$

(d) $\displaystyle \frac{\epsilon_0 a^2}{d}(1-\frac{3\alpha a}{2d})$

Click to See Solution :
Ans: (c)

Sol: Let us consider a small element of thickness dx at a distance x from left Capacitance of small element $\displaystyle dC = \frac{\epsilon_0 a dx}{d + x \alpha}$

$\displaystyle C = \int_{0}^{a} \frac{\epsilon_0 a dx}{d + x \alpha}$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ln(1+\frac{x \alpha}{d})]_{0}^{a}$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ln(1+\frac{a \alpha}{d}) – ln1]$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}ln(1+\frac{a \alpha}{d})$

Applying $ln(1+y) = y – \frac{y^2}{2} + \frac{y^3}{3} – …..$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}[ \frac{a \alpha}{d} – \frac{a^2 \alpha^2}{2d^2}]$

$\displaystyle C = \frac{\epsilon_0 a}{\alpha}\times \frac{a \alpha}{d}[ 1 – \frac{a \alpha}{2d}]$

$\displaystyle C = \frac{\epsilon_0 a^2}{d}(1-\frac{\alpha a}{2d})$

## In finding the electric field using Gauss law the formula E = q_enc/∈0 A is applicable ….

Q: In finding the electric field using Gauss law the formula $|\vec{E}| = \frac{q_{enc}}{\epsilon_0 |A|}$ is applicable . In the formula ∈0 is permittivity of free space , A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface . This equation can be used in which of the following situation ?

(a) For any choice of Gaussian surface

(b) Only when the Gaussian surface is an equipotential surface

(c) Only when the Gaussian surface is an equipotential surface and $\vec{E}$ is constant on the surface

(d) Only when $\vec{E}$ = constant on the surface

Click to See Solution :
Ans: (c)
Sol: The magnitude of electric field should be constant and surface should be equipotential surface .