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1

AIEEE 2012

MCQ (Single Correct Answer)
This question has statement- $$1$$ and statement- $$2.$$ Of the four choices given after the statements, choose the one that best describe the two statements.
An insulating solid sphere of radius $$R$$ has a uniformly positive charge density $$\rho $$. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero.

Statement- $$1:$$ When a charge $$q$$ is take from the centre of the surface of the sphere its potential energy changes by $${{q\rho } \over {3{\varepsilon _0}}}$$
Statement- $$2:$$ The electric field at a distance $$r\left( {r < R} \right)$$ from the center of the sphere is $${{\rho r} \over {3{\varepsilon _0}}}.$$

A
Statement- $$1$$ is true, Statement- $$2$$ is true; Statement- $$2$$ is not the correct explanation of Statement- $$1$$.
B
Statement $$1$$ is true, Statement $$2$$ is false.
C
Statement $$1$$ is false, Statement $$2$$ is true.
D
Statement- $$1$$ is true, Statement- $$2$$ is true; Statement- $$2$$ is the correct explanation of Statement- $$1$$.

Explanation

The electric field inside a uniformly charged sphere is

= $${{\rho .r} \over {3{ \in _0}}}$$

The electric potential inside a uniformly charged sphere

$$ = {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - {{{r^2}} \over {{R^2}}}} \right]$$

$$\therefore$$ Potential difference between center and surface

$$ = {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - 2} \right] = {{\rho {R^2}} \over {6{ \in _0}}}$$

$$\Delta U = {{q\rho {R^2}} \over {6{ \in _0}}}$$
2

AIEEE 2011

MCQ (Single Correct Answer)
Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d\left( {d < < 1} \right)$$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $$v$$. Then as a function of distance $$x$$ between them,
A
$$v\, \propto \,{x^{ - 1}}$$
B
$$y\, \propto \,{x^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
C
$$v\, \propto \,x$$
D
$$v\, \propto \,{x^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

Explanation

At any instant

$$T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$T\sin \theta = {F_e}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$ \Rightarrow {{\sin \theta } \over {\cos \theta }} = {{{F_e}} \over {mg}} \Rightarrow {F_e} = mg\,\tan \theta $$

$$ \Rightarrow {{k{q^2}} \over {{x^2}}} = mg\,\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta $$

$$\sin \theta = {\textstyle{x \over {2l}}}$$

For small $$\theta ,\,\sin \theta \approx \tan \theta $$

$$\therefore$$ $${q^2} \propto {x^3}$$



$$ \Rightarrow q{{dq} \over {dt}} \propto {x^2}{{dx} \over {dt}}$$

$$\therefore$$ $${{dq} \over {dt}} = const.$$

$$\therefore$$ $$q \propto {x^2}.v \Rightarrow {x^{3/2}}\alpha {x^2}.v\,\,$$ $$\,\,\,\,\,$$ $$\left[ {\,\,} \right.$$ as $$\left. {{q^2} \propto {x^3}\,\,} \right]$$

$$ \Rightarrow v \propto {x^{ - 1/2}}$$
3

AIEEE 2011

MCQ (Single Correct Answer)
The electrostatic potential inside a charged spherical ball is given by $$\phi = a{r^2} + b$$ where $$r$$ is the distance from the center and $$a,b$$ are constants. Then the charge density inside the ball is:
A
$$ - 6a{\varepsilon _0}r$$
B
$$ - 24\pi a{\varepsilon _0}$$
C
$$ - 6a{\varepsilon _0}$$
D
$$ - 24\pi {\varepsilon _0}r$$

Explanation

Electric field

$$E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

By Gauss's theorem

$$E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$\left( i \right)$$ and $$\left( ii \right),$$

$$q = - 8\pi {\varepsilon _0}a{r^3}$$

$$ \Rightarrow dq = - 24\pi {\varepsilon _0}ar{}^2dr$$

Charge density, $$\rho = {{dq} \over {4\pi {r^2}dr}} = - 6{\varepsilon _0}a$$
4

AIEEE 2010

MCQ (Single Correct Answer)
A thin semi-circular ring of radius $$r$$ has a positive charges $$q$$ distributed uniformly over it. The net field $$\overrightarrow E $$ at the center $$O$$ is
A
$${q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$
B
$$ - {q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$
C
$$ - {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$
D
$$ {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Explanation



Let us consider a differential element $$dl.$$ charge on this element.

$$dq = \left( {{q \over {\pi r}}} \right)dl$$

$$ = {q \over {\pi r}}\left( {rd\theta } \right)\,\,\,\,\,$$ (as $$dl = rd\theta $$)

$$ = \left( {{q \over \pi }} \right)d\theta $$

Electric field at $$O$$ due to $$dq$$ is

$$dE = {1 \over {4\pi { \in _0}}}.{{dq} \over {{r^2}}}$$

$$ = {1 \over {4\pi { \in _0}}}.{q \over {\pi {r^2}}}d\theta $$

The component $$dE\cos \theta $$ will be counter balanced by another element on left portion.

Hence resultant field at $$O$$ is the resultant of the component $$dE\sin \theta $$ only.

$$\therefore$$ $$E = \int {dE\sin \theta = \int\limits_0^\pi {{q \over {4{\pi ^2}{r^2}{ \in _0}}}} } \sin \theta d\theta $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi $$

$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right)$$

$$ = {q \over {2{\pi ^2}{r^2}{ \in _0}}}$$

The directions of $$E$$ is towards negative $$y$$-axis.

$$\therefore$$ $$\overrightarrow E = - {q \over {2{\pi ^2}{r^2}{ \in _0}}}\widehat j$$

Questions Asked from Electrostatics

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