Practicing Success
Supposing that the earth has a surface charge density of 1 electron/m2; calculate earth's potential |
− 0.115 Volt − 0.110 Volt − 0.105 Volt − 0.112 Volt |
− 0.115 Volt |
If R be the radius and σ the surface charge density of earth, then the total charge q on its surface is given by Q = 4πR2σ (i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated as its centre. Thus $V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{4 \pi R^2 \sigma}{R}=\frac{R \sigma}{\varepsilon_0}$ Substituting the given value: $V=\frac{6.4 \times 10^6 \times\left(-1.6 \times 10^{-19}\right)}{8.9 \times 10^{-12}}$ = − 0.115 N-m/C = −0.115 J/C = − 0.115Volt. |