## A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance…..

Q. A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball’s centre as ρ = ρ0 (1 − r/R). Where ρ0 is a constant. Assume ε  as the permittivity of space. The magnitude of electric field as a function of the distance r inside the sphere is given by

(a) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}-\frac{r^2}{4R}]$

(b) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]$

(c) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}+ \frac{r^2}{4R}]$

(d) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]$

Ans: (a)

Sol: charge inside a sphere of radius r (r < R)

$\displaystyle q = \int_{0}^{r} \rho dv$

$\displaystyle = \int_{0}^{r} \rho_0 (1-\frac{r}{R})4\pi r^2 dr$

$\displaystyle q = 4\pi \rho_0 \int_{0}^{r}(r^2 -\frac{r^3}{R})$

$\displaystyle q = 4\pi \rho_0 (\frac{r^3}{3}-\frac{r^4}{4R})$

$\displaystyle E = \frac{1}{4\pi \epsilon}\frac{q}{r^2}$

$\displaystyle E = \frac{1}{4\pi \epsilon}.4\pi \rho_0 (\frac{r}{3}-\frac{r^2}{4R})$

$\displaystyle \frac{\rho_0}{\epsilon}(\frac{r}{3}-\frac{r^2}{4R})$

## Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

Q. Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

(a) $\displaystyle \frac{\rho R^2}{6\epsilon_0}$

(b) $\displaystyle \frac{\rho R^2}{4\epsilon_0}$

(c) $\displaystyle \frac{\rho R^2}{3\epsilon_0}$

(d) $\displaystyle \frac{\rho R^2}{2\epsilon_0}$

Ans: (a)
Charge on Sphere is

$\displaystyle q = \frac{4}{3}\pi R^3 \rho$

Electric Potential at the centre

$\displaystyle V_o = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R}$

Electric Potential at the Surface

$\displaystyle V_P = \frac{1}{4\pi \epsilon_0}\frac{q}{R}$

Potential difference between centre and surface

$\displaystyle = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R} - \frac{1}{4\pi \epsilon_0}\frac{q}{R}$

$\displaystyle = \frac{1}{4\pi \epsilon_0}\frac{q}{2R}$

$\displaystyle = \frac{1}{4\pi \epsilon_0 (2R)}(\frac{4}{3}\pi R^3 \rho)$

$\displaystyle = \frac{\rho R^2}{6\epsilon_0}$

## Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line…..

Q. Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is (x >> a)

(a)Zero

(b) $\displaystyle \frac{Qa}{\pi\epsilon_0 x^3 }$

(c) $\displaystyle \frac{2Qa}{\pi\epsilon_0 x^3 }$

(d) $\displaystyle \frac{\sqrt3 Qa}{\pi\epsilon_0 x^3 }$

Ans: (b)

Electric Field due to dipole at a point on the axis passing through O is

$\displaystyle E = \frac{p}{4\pi \epsilon_0 x^3}$

Where p = Q × 2a

Net field is
$\displaystyle E_{net} = 2E = \frac{2 \times Q \times 2 a}{4\pi \epsilon_0 x^3}$

$\displaystyle = \frac{Q a}{\pi\epsilon_0 x^3 }$

## Two thin wire rings each having radius R are placed at a distance d apart with their axes coinciding…..

Q. Two thin wire rings each having radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +Q and -Q The potential difference between the centres of the two rings is

(a)Zero

(b) $\displaystyle \frac{Q}{4\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]$

(c) $\displaystyle \frac{Q}{4\pi\epsilon_0 d^2}$

(d) $\displaystyle \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]$

Ans: (d)

Sol:
$\displaystyle V_1 = \frac{-Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{Q}{4\pi \epsilon_0 R}$

$\displaystyle V_2 = \frac{Q}{4\pi \epsilon_0 \sqrt{R^2 + d^2}} + \frac{-Q}{4\pi \epsilon_0 R}$

$\displaystyle V_1 -V_2 = \frac{Q}{2\pi\epsilon_0 } [\frac{1}{R}-\frac{1}{\sqrt{ R^2 +d^2}}]$

## An infinite line of charge λ per unit length is placed along the y-axis. The work done in moving a charge…..

Q. An infinite line of charge λ per unit length is placed along the y-axis. The work done in moving a charge q from A(a , 0) to B(2a , 0) is

(a) $\displaystyle \frac{q\lambda}{2\pi\epsilon_0} ln2$

(b) $\displaystyle \frac{q\lambda}{2\pi\epsilon_0} ln \frac{1}{2}$

(c) $\displaystyle \frac{q\lambda}{4\pi\epsilon_0} ln\sqrt2$

(d) $\displaystyle \frac{q\lambda}{4\pi\epsilon_0} ln2$

Ans: (b)
Sol:
$\displaystyle W= -\int_{A}^{B}\vec{F}.\vec{dr}$

$\displaystyle = -q \int_{a}^{2a}Edr$

$\displaystyle = -q\int_{a}^{2a}\frac{\lambda}{2\pi \epsilon_0 r}dr$