Question:
Two non-conducting solid spheres of radii $R$ and $2R$ , having uniform volume charge densities $\rho _1 $ and $ \rho _2 $ respectively, touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere, along the line joining the centre of the spheres, is zero. The ratio $\rho _1 \, / \, \rho _2 $ can be
Two non-conducting solid spheres of radii $R$ and $2R$ , having uniform volume charge densities $\rho _1 $ and $ \rho _2 $ respectively, touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere, along the line joining the centre of the spheres, is zero. The ratio $\rho _1 \, / \, \rho _2 $ can be
Updated On: Jun 14, 2022
- $-4$
- $\frac {-32}{25}$
- $\frac {32}{25} $
- $4$
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The Correct Option is D
Solution and Explanation
At point $P_{1}, \frac{1}{4 \pi \varepsilon_{0}} \frac{\rho_{1}(4 / 3) \pi R^{3}}{4 R^{2}}=\frac{\rho_{2} R}{3 \varepsilon_{0}}$
$\frac{\rho_{1} R}{12}=\frac{\rho_{2} R}{3}$
$\frac{\rho_{1}}{\rho_{2}}=4$
At point $P _{2}$
$\frac{\rho_{1}(4 / 3) \pi R ^{3}}{(2 R )^{2}}+\frac{\rho_{2}(4 / 3) \pi 8 R ^{3}}{(5 R )^{2}}=0$
$\therefore \frac{\rho_{1}}{\rho_{2}}=-\frac{32}{25}$
$\frac{\rho_{1} R}{12}=\frac{\rho_{2} R}{3}$
$\frac{\rho_{1}}{\rho_{2}}=4$
At point $P _{2}$
$\frac{\rho_{1}(4 / 3) \pi R ^{3}}{(2 R )^{2}}+\frac{\rho_{2}(4 / 3) \pi 8 R ^{3}}{(5 R )^{2}}=0$
$\therefore \frac{\rho_{1}}{\rho_{2}}=-\frac{32}{25}$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).