Chapter 16, Problem 65GP

To determine

To Find: Amount of charge that can be stored by pea before spontaneous discharge

Answer to Problem 65GP

  $5\times {10}^9\text{C}$

Explanation of Solution

Given:

Break down electric field strength for

  ${\mathrm{E}}_{\mathrm{a}\mathrm{i}\mathrm{r}}=3\times {10}^6\text{N/C}$

Diameter of given pea is 0.75 cm.

Formula used:

Electric field on the surface of pea is given by

  $\mathrm{E}=\frac{\mathrm{K}\mathrm{Q}}{{\mathrm{r}}^2}$

  $\Rightarrow \mathrm{Q}=\frac{\mathrm{E}{\mathrm{r}}^2}{\mathrm{K}}$

Calculation:

The pea will discharge when the electric field at the surface exceeds the breakdown field.

Because the charge on the pea as appoint charge at the center,

  $\mathrm{E}=\frac{\mathrm{K}\mathrm{Q}}{{\mathrm{r}}^2}$

  $3\times {10}^6\text{N}/\text{C}=\frac{\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\mathrm{Q}}{{\left(0.375\times {10}^{-2}\text{m}\right)}^2}$

This gives $\mathrm{Q}=5\times {10}^{-9}\text{C}$ .