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1

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.

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

AIEEE 2010

MCQ (Single Correct Answer)
Let there be a spherically symmetric charge distribution with charge density varying as $$\rho \left( r \right) = {\rho _0}\left( {{5 \over 4} - {r \over R}} \right)$$ upto $$r=R,$$ and $$\rho \left( r \right) = 0$$ for $$r>R,$$ where $$r$$ is the distance from the erigin. The electric field at a distance $$r\left( {r < R} \right)$$ from the origin is given by
A
$${{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$
B
$${{4\pi {\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$
C
$${{4{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$$
D
$${{{\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$$

Explanation



Let us consider a spherical shell of radius $$x$$ and thickness $$dx.$$

Charge on this shell

$$dq = \rho .4{\pi ^2}dx = {\rho _0}\left( {{5 \over 4} - {x \over R}} \right).4\pi {x^2}dx$$

$$\therefore$$ Total charge in the spherical region from center to $$r$$ $$\left( {r < R} \right)$$ is

$$q = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {{5 \over 4} - {x \over R}} \right)} } {x^2}dx$$

$$ = 4\pi {\rho _0}\left[ {{5 \over 4}.{{{r^3}} \over 3} - {1 \over R}.{{{r^4}} \over 4}} \right]$$

$$ = \pi {\rho _0}{r^3}\left( {{5 \over 3} - {r \over R}} \right)$$

$$\therefore$$ Electric field at $$r,$$ $$E = {1 \over {4\pi { \in _0}}}.{q \over {{r^2}}}$$

$$ = {1 \over {4\pi { \in _0}}}.{{\pi {\rho _0}{r^3}} \over {{r^2}}}\left( {{5 \over 3} - {r \over R}} \right)$$

$$ = {{{\rho _0}r} \over {4{ \in _0}}}\left( {{5 \over 3} - {r \over R}} \right)$$
3

AIEEE 2009

MCQ (Single Correct Answer)
A charge $$Q$$ is placed at each of the opposite corners of a square. A charge $$q$$ is placed at each of the other two corners. If the net electrical force on $$Q$$ is zero, then $$Q/q$$ equals:
A
$$-1$$
B
$$1$$
C
$$ - {1 \over {\sqrt 2 }}$$
D
$$ - 2\sqrt 2 $$

Explanation



Let $$F$$ be the force between $$Q$$ and $$Q.$$ The force between $$q$$ and $$Q$$ should be attractive for net force on $$Q$$ to be zero. Let $$F'$$ be the force between $$Q$$ and $$q.$$ For equilibrium

$$\sqrt 2 F' = - F$$

$$\sqrt 2 \times k{{Qq} \over {{\ell ^2}}} = - k{{{Q^2}} \over {{{\left( {\sqrt 2 \ell } \right)}^2}}}$$

$$ \Rightarrow {Q \over q} = - 2\sqrt 2 $$
4

AIEEE 2009

MCQ (Single Correct Answer)
Two points $$P$$ and $$Q$$ are maintained at the potentials of $$10$$ $$V$$ and $$-4$$ $$V$$, respectively. The work done in moving $$100$$ electrons from $$P$$ to $$Q$$ is :
A
$$9.60 \times {10^{ - 17}}J$$
B
$$ - 2.24 \times {10^{ - 16}}J$$
C
$$ 2.24 \times {10^{ - 16}}J$$
D
$$- 9.60 \times {10^{ - 17}}J$$

Explanation

$$ {{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)$$

$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$

$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$

$$ = + 2.24 \times {10^{ - 16}}J$$

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