Suppose that the earth has a surface charge density of 1 electron per \( \mathrm{m^{2}} \). Calculate the electric field just outside the earth’s surface. The electronic charge is \( -1.6\times10^{-19}\,\mathrm{C} \) and the earth’s radius is \( 6.4\times10^{6}\,\mathrm{m} \) \( \left(\varepsilon_0=8.9\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}\right) \).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -1.8\times10^{-8}\,\mathrm{N/C} \)
The earth is treated as a charged conductor. The field just outside its surface is normal to the surface, and its value is \( E=\sigma/\varepsilon_0 \). Here \( \sigma \) is the surface charge density, that is, the charge on each square metre of the surface. The radius of the earth is not needed.
One electron on each square metre gives:
\( \sigma=-1.6\times10^{-19}\,\mathrm{C\,m^{-2}} \)
The field just outside the surface is:
\( \begin{aligned} E &= \frac{\sigma}{\varepsilon_0} \\ &= \frac{-1.6\times10^{-19}\,\mathrm{C\,m^{-2}}}{8.9\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}} \\ &\approx -1.8\times10^{-8}\,\mathrm{N/C} \end{aligned} \)
The negative sign shows that the field points inward, towards the centre of the earth.
NCERT (2026-27):
Electrostatics of Conductors – “5. Electric field at the surface of a charged conductor
\( \displaystyle \mathbf{E}=\frac{\sigma}{\varepsilon_0}\,\hat{\mathbf{n}} \)\( \qquad (2.35) \)
where σ is the surface charge density and \( \hat{\mathbf{n}} \) is a unit vector normal to the surface in the outward direction.”
Electrostatics of Conductors – “… Including the fact that electric field is normal to the surface, we get the vector relation, Eq. (2.35), which is true for both signs of σ. For σ > 0, electric field is normal to the surface outward; for σ < 0, electric field is normal to the surface inward.”