91Ó°ÊÓ

Magnitude of electric field\(E = {\bf{1}} \times {\bf{1}}{{\bf{0}}^{\bf{6}}}{\rm{ }}{\bf{N}}/{\bf{C}}\)

Distance \(r = 25\,{\rm{m}}\)

Each location in space where a charge exists in any form can be considered to have an electric field attached to it. It is expressed as,

\(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left| q \right|}}{{{r^2}}}\) …(i)

Where, \({\varepsilon _0}\) is permittivity of free space, \(q\) is the charge surface and \(r\) is the distance

Rearranging the equation (i)

\(\left| q \right| = 4\pi {\varepsilon _0}{r^2}E\)

Substitute \(4\pi {\varepsilon _0} = \frac{1}{{9 \times {{10}^9}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}}}\) in above equation

\(\begin{aligned}\left| q \right| = \frac{1}{{9 \times {{10}^9}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}}} \times {\left( {25\,{\rm{m}}} \right)^2} \times \left( {{\bf{1}} \times {\bf{1}}{{\bf{0}}^{\bf{6}}}{\rm{ }}{\bf{N}}/{\bf{C}}} \right)\\ = 0.0694{\rm{C}}\,\\ \approx 0.07{\rm{C}}\end{aligned}\)

The polarity of charge will be negative because electric field intends to repel the electrons.

So net charge on sphere is \( - 0.07{\rm{C}}\)

Hence option (a) is correct.