Question

Figure shows a uniformly charged hemisphere of radius R. It has a volume charge density ρ. If the electric field at a point 2R, above its centre is E, then what is the electric field at the point 2R below its centre?

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Answer 1

let us assume one more hemisphere over this one with same size and charge such that a solid sphere is formed.

Therefore, electric field due to this sphere at a distance 2R is given by

E'=

$\frac{ \rho \times\frac{4}{3} \pi \times {r}^{3} }{4\pi \epsilon {(2r)}^{2} } = \frac{ \rho r}{12\epsilon}$

and,

E'= electric field due to the upper hemi-sphere + electric field due to the lower hemi-sphere

let electric field due to the lower hemi-sphere = electric field due to the hemisphere at a distance 2R below its centre = h(say)

therefore,

$\frac{ \rho r}{12 \epsilon} = e + h \\ so \: required \: electric \: field \: is \: given \: by \\ h = \frac{ \rho r}{12 \epsilon} - e$

and so option (b) is correct.

Hope it helps（＾＿－）≡★

Answer 2

Answer:

The electric field at the point $2R$ below its center is $\bold{h=\frac{\rho r}{12\epsilon} -e}$

Explanation:

Let us assume that one more hemisphere over this one with the same size and charge such that a solid sphere is formed.

Therefore, the electric field due to this sphere at a distance $2R$ is given by

$E'=\frac{\rho\times\frac{4}{3}\times r^{3}}{4\pi\epsilon(2r^2)} \\=\frac{\rho r}{12\epsilon }$

Now let electric field due to lower hemisphere = electric field due to the hemisphere at a distance $2R$ below its center = h

Therefore. $\frac{\rho r}{12\epsilon }=e+h$

So, the required electric field is given by $\bold{h=\frac{\rho r}{12\epsilon} -e}$

Hence, option (b) is correct.