Question

An electric field of approximately $\mathbf{100}\mathbf{\text{\hspace{0.17em}V/m}}$is often observed near the surface of Earth. If this were the field over the entire surface, what would be the electric potential of a point on the surface? (Set$\mathbf{\text{V=0}}$ at infinity.)

Short answer

$\mathrm{V}=6.4\times {10}^8\text{\hspace{0.17em}V}$

Step-by-step solution

Given

Intensity of the electric field$\mathrm{E}=100\text{\hspace{0.17em}V/m}$
Radius of the earth's spherical surface,${\mathrm{R}}_{\mathrm{c}}=6400\text{km}$

Understanding the concept

Use the relation between E and V.

Electric potential at the surface of the Earth,$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{R}}_{\mathrm{c}}}$
Electric intensity at the surface of earth,$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left(\frac{\mathrm{q}}{{\mathrm{R}}_{\mathrm{e}}^2}\right)=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{e}}}$

Calculate the electric potential  

Electric potential,

$\begin{array}{c}\mathrm{V}={\mathrm{ER}}_{\mathrm{e}}\\ \mathrm{V}=(100\text{\hspace{0.17em}V/m})(6400\times {10}^3\text{\hspace{0.17em}m})\\ \mathrm{V}=6.4\times {10}^8\text{\hspace{0.17em}V}\end{array}$

Hence, the electric potential of a point on the surface is $\mathrm{V}=6.4\times {10}^8\text{\hspace{0.17em}V}$