JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 1)
A solid ball of radius R has a charge density $$\rho $$
given by $$\rho $$ = $$\rho $$o (1 $$-$$ $${\raise0.5ex\hbox{$\scriptstyle r$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle R$}}$$) for 0 $$ \le $$ r $$ \le $$ R. The electric field outside the ball is :
given by $$\rho $$ = $$\rho $$o (1 $$-$$ $${\raise0.5ex\hbox{$\scriptstyle r$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle R$}}$$) for 0 $$ \le $$ r $$ \le $$ R. The electric field outside the ball is :
$${{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}$$
$${{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}$$
$${{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}$$
$${{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}$$
Explanation
Electric field outside the ball is given by
E = $${1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$$ .............(i)
Now, dq = $$\rho $$dV = $$\rho $$(4$$\pi $$r2)dr
$$ \therefore $$ q = $$\int {dq = \int\limits_0^R {{\rho _0}\left( {1 - {r \over R}} \right)\left( {4\pi {r^2}} \right)dr} } $$
= $${\left( {4\pi {\rho _0}} \right)\left[ {{{{r^3}} \over 3} - {1 \over R} \times {{{r^4}} \over 4}} \right]_0^R}$$
= $${\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over 3} - {{{R^3}} \over 4}} \right)}$$
$$ \Rightarrow $$ q = $${\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over {12}}} \right)}$$ ......(ii)
From eqns. (i) and (ii), E = $${{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}$$
E = $${1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$$ .............(i)
Now, dq = $$\rho $$dV = $$\rho $$(4$$\pi $$r2)dr
$$ \therefore $$ q = $$\int {dq = \int\limits_0^R {{\rho _0}\left( {1 - {r \over R}} \right)\left( {4\pi {r^2}} \right)dr} } $$
= $${\left( {4\pi {\rho _0}} \right)\left[ {{{{r^3}} \over 3} - {1 \over R} \times {{{r^4}} \over 4}} \right]_0^R}$$
= $${\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over 3} - {{{R^3}} \over 4}} \right)}$$
$$ \Rightarrow $$ q = $${\left( {4\pi {\rho _0}} \right)\left( {{{{R^3}} \over {12}}} \right)}$$ ......(ii)
From eqns. (i) and (ii), E = $${{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}$$
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