Chapter 16, Problem 26P

To determine

To find: the magnitude and direction of the electric field

Answer to Problem 26P

Direction of the electric field is Direct outward or up

Magnitude of the electric field is $\mathrm{E}=7.425\times {10}^6\text{N}/\text{C}$

Explanation of Solution

Given:

  $\mathrm{q}=33\times {10}^{-6}\mathrm{C},\mathrm{r}=20\text{cm}$

Formula used:

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

Calculation:

Source charge $\mathrm{q}=33\times {10}^{-6}\text{C}$

Distance from the charge $\mathrm{r}=20\text{cm}$

  $=0.2\text{m}$

Magnitude of electric field $\mathrm{E}=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}$

  $\mathrm{E}=\frac{\left(9\times {10}^9\right)\left(33\times {10}^{-6}\right)}{{0.2}^2}\text{N/C}$

  $\mathrm{E}=7.425\times {10}^6\text{N}/\text{C}$

Direction of the electric field is direct outward or up