JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 14)
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by $$\rho $$(r) = kr, where
r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite
points, at equal distance, $$a$$ from the centre. If A and B do not experience any force, then :
$$a = {8^{ - 1/4}}R$$
$$a = {2^{ - 1/4}}R$$
$$a = {{3R} \over {{2^{1/4}}}}$$
$$a = {R \over {\sqrt 3 }}$$
Explanation
Total charge = 2Q
Charging density $$\rho $$ = kr
Radius = R_12th_April_Evening_Slot_en_14_1.png)
Charge enclosed in the sphere,
qin = $$\int\limits_0^R {\rho dV} $$
$$ \Rightarrow $$ 2Q = $$\int\limits_0^R {kr4\pi {r^2}dr} $$
$$ \Rightarrow $$ 2Q = $$k4\pi \int\limits_0^R {{r^3}dr} $$
$$ \Rightarrow $$ 2Q = $$k4\pi {{{R^4}} \over 4}$$
$$ \Rightarrow $$ k = $${{2Q} \over {\pi {R^4}}}$$ ........... (1)
Force on charge at A will be due to charge at B and due to force applied by the charge in sphere._12th_April_Evening_Slot_en_14_2.png)
Here Fsphere = EQ
Using Gauss law, we can find electric field at point A due to sphere,
∮ $$\overrightarrow E .d\overrightarrow A $$ = $${{{q_{in}}} \over {{ \in _0}}}$$
$$ \Rightarrow $$ $$E\left( {4\pi {a^2}} \right)$$ = $${{\int\limits_0^a {\rho dV} } \over {{ \in _0}}}$$
$$ \Rightarrow $$ $$E\left( {4\pi {a^2}} \right)$$ = $${{k4\pi {{{a^4}} \over 4}} \over {{ \in _0}}}$$
$$ \Rightarrow $$ E = $${{k{a^2}} \over {4{ \in _0}}}$$
As on charge A net force is zero then,
FAB = Fsphere
$$ \Rightarrow $$ $${{Q \times Q} \over {4\pi { \in _0}{{\left( {2a} \right)}^2}}}$$ = $${{k{a^2}} \over {4{ \in _0}}}$$ $$ \times $$ Q
$$ \Rightarrow $$ $${Q \over {4\pi {a^2}}} = k{a^2}$$
$$ \Rightarrow $$ $${Q \over {4\pi {a^2}}} = {{2Q} \over {\pi {R^4}}}{a^2}$$ [ from equation (1)]
$$ \Rightarrow $$ $$8{a^4} = {R^4}$$
$$ \Rightarrow $$ $$a = {8^{ - 1/4}}R$$
Charging density $$\rho $$ = kr
Radius = R
Charge enclosed in the sphere,
qin = $$\int\limits_0^R {\rho dV} $$
$$ \Rightarrow $$ 2Q = $$\int\limits_0^R {kr4\pi {r^2}dr} $$
$$ \Rightarrow $$ 2Q = $$k4\pi \int\limits_0^R {{r^3}dr} $$
$$ \Rightarrow $$ 2Q = $$k4\pi {{{R^4}} \over 4}$$
$$ \Rightarrow $$ k = $${{2Q} \over {\pi {R^4}}}$$ ........... (1)
Force on charge at A will be due to charge at B and due to force applied by the charge in sphere.
Here Fsphere = EQ
Using Gauss law, we can find electric field at point A due to sphere,
∮ $$\overrightarrow E .d\overrightarrow A $$ = $${{{q_{in}}} \over {{ \in _0}}}$$
$$ \Rightarrow $$ $$E\left( {4\pi {a^2}} \right)$$ = $${{\int\limits_0^a {\rho dV} } \over {{ \in _0}}}$$
$$ \Rightarrow $$ $$E\left( {4\pi {a^2}} \right)$$ = $${{k4\pi {{{a^4}} \over 4}} \over {{ \in _0}}}$$
$$ \Rightarrow $$ E = $${{k{a^2}} \over {4{ \in _0}}}$$
As on charge A net force is zero then,
FAB = Fsphere
$$ \Rightarrow $$ $${{Q \times Q} \over {4\pi { \in _0}{{\left( {2a} \right)}^2}}}$$ = $${{k{a^2}} \over {4{ \in _0}}}$$ $$ \times $$ Q
$$ \Rightarrow $$ $${Q \over {4\pi {a^2}}} = k{a^2}$$
$$ \Rightarrow $$ $${Q \over {4\pi {a^2}}} = {{2Q} \over {\pi {R^4}}}{a^2}$$ [ from equation (1)]
$$ \Rightarrow $$ $$8{a^4} = {R^4}$$
$$ \Rightarrow $$ $$a = {8^{ - 1/4}}R$$
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