Chapter 16, Problem 55GP

To determine

To Find: Magnitude of electric charge on Earth and state whether it is positive or negative

Answer to Problem 55GP

  $6.82\times {10}^5\text{C}$

It is negative.

Explanation of Solution

Given:

The electric field E =150N/ C

The radius of Earth r = 6400 km

  $=6.4\times {10}^6\text{m}$

Formula used:

Electric field is given by

  $\begin{array}{l}\mathrm{E}=\frac{\mathrm{K}\mathrm{Q}}{{\mathrm{r}}^2}\\ \mathrm{Q}=\frac{\mathrm{E}{\mathrm{r}}^2}{\mathrm{K}}\end{array}$

Calculations:

Let the charge on the Earth be = q

So ,

Electric field , $\mathrm{E}=\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{1}{{\mathrm{R}}^2}$

  $\begin{array}{l}150=\left(9.0\times {10}^9\right)\frac{\mathrm{q}}{{\left(6.4\times {10}^6\right)}^2}\\ \mathrm{q}=\frac{150\times {\left(6.4\times {10}^6\right)}^2}{9.0\times {10}^9}\text{C}\\ \mathrm{q}=6.82\times {10}^5\text{C}\end{array}$

Since the electric field is pointing towards the Earth’s center, the charge must be negative.

Conclusions:

Charge on Earth is $6.82\times {10}^{\text{5}}\text{C}$

The charge is negative.