Question 613: Electrostats & Capacitance

Question

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 :

Accepted Answer

Electric field outside the ball is given by
E = $${1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$$ .............(i)
Now, dq = $$ho $$dV = $$ho $$(4$$\pi $$r
2
)dr
$$ 	herefore $$ q = $$\int {dq = \int\limits_0^R {{ho _0}\left( {1 - {r \over R}} ight)\left( {4\pi {r^2}} ight)dr} } $$
= $${\left( {4\pi {ho _0}} ight)\left[ {{{{r^3}} \over 3} - {1 \over R} 	imes {{{r^4}} \over 4}} ight]_0^R}$$
= $${\left( {4\pi {ho _0}} ight)\left( {{{{R^3}} \over 3} - {{{R^3}} \over 4}} ight)}$$
$$ \Rightarrow $$ q = $${\left( {4\pi {ho _0}} ight)\left( {{{{R^3}} \over {12}}} ight)}$$  ......(ii)
From eqns. (i) and (ii), E = $${{{ho _o}{R^3}} \over {12{ \in _o}{r^2}}}$$

Answer

$${{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}$$

Answer

$${{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}$$

Answer

$${{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}$$

Answer

$${{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}$$

Electrostats & Capacitance

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 :

Choose the correct answer: