Question

the surface density of conducting sphere is 7.7 into 10 to the power minus 10 coulomb per metre square and electric field intensity at a distance of 4 m from the centre of the sphere is 200 into 10 to the power minus 2 volt metre find the radius of the sphere assuming the sphere to be vacuum​

Answer 1

$\mathfrak{\red{ \underline{ \underline{answer}}}} \implies \: \\ →26.4 m \\ \\ \bf{ \underline{explanation}} : -$

Given:-

$distance \: (d) = 4 \: m \\ charge \: density \: \: ( \sigma) = 7.7 \times {10}^{ - 10} \frac{c}{ {m}^{2} } \\ electric \: field \: intensity \: (E )= 200 \times {10}^{ - 2} \: \: v \: m$

To find :-

Radius of sphere =?

Let permittivity of free space is g.

$\underline{\bf{soltion}} : - \\ \\ let \: radius \: of \: sphere \: is \: r \: \\ \\ we \: know \: that \\ \\ \implies \: E = \frac{ \sigma \: {r}^{2} }{g {d}^{2} } \\ \\ \implies \: 200 \times {10}^{ - 2} = \frac{7.7 \times {10}^{ - 10 } \times 4 \times 4}{8.85 \times {10}^{ - 12} \times {r}^{2} } \\ \\ \implies \: {r}^{2} = \frac{7.7 \times 16 \times {10}^{ - 10} }{8.85 \times 200 \times {10}^{ - 14} } \\ \\ \implies \: {r}^{2} = \frac{1232 \times {10}^{3} }{885 \times 2} \\ \\ \implies \: {r}^{2} = \frac{1232}{1770} \times {10}^{3} \\ \\ \implies \: r = 2.64 \times 10 \\ \\ \: \implies \: r = 26.4 \: m \: (aprrox)$