Question

The electric field at $20$ cm from the centre of a uniform charged non - conducting sphere of radius $10$ cm is $E$. Then at a distance of $5$ cm from the centre it will be :

Answer 1

For a non conducting sphere containing a distributed charge $Q$ and of rations $R$ the electric field in given at a distance ${}^{\prime }{r}^{\prime }$

$E=\{\begin{array}{c}\frac{KQ}{r^2},whenr>R\left(outside\right)\\ \frac{pr}{{3t}_o},whenr>R\left(inside\right)andherep=\frac{Q}{v}\left[chargedensity\right]\end{array}$

at a distance $r=20cm$, field is $E(r>R)$ where $R=10cm$ (Radius of sphere)

$E=\frac{KQ}{2O^2}\Rightarrow Q=\frac{400E}{K}=1600\pi {ϵ}_{0}E$ $(asK=\frac{1}{4\pi {ϵ}_0})$

now volume of sphere $V=\frac{4}{3}\pi R^3$

So at a distance $r=5cm,(r<R),$ inside sphere :-

$E^{{}^{\prime }}=\frac{pr}{3{ϵ}_0}=\frac{Qr}{v3{ϵ}_0}$

$=\frac{1600\pi {ϵ}_{0}E\times 5}{\frac{4}{3}\pi \times {10}^3\times 3\times {ϵ}_0}=\frac{2000E}{1000}=2E$

$\therefore E^{{}^{\prime }}=2E$ (Ans)