Question

A -12 nC charge is located at (x, y) =(1 CM, 0 CM) what are the electric fields at the positions (x, y) =(5,0) CM (-5,0)cm and (0,5)cm write each electric field vector in component form

Answer 1

Answer on Question #66219-Physics - Field Theory

A -12 nC charge is located at (x, y) = (1 CM, 0 CM) what are the electric fields at the positions (x, y) = (5,0) CM (-5,0)cm and (0,5)cm write each electric field vector in component form.

Solution

The charge position $\mathbf{r} = (1,0)$.

The electric field at the point $\mathbf{r}'$ is

$$\mathbf{E} = k \frac{q}{|\mathbf{r} - \mathbf{r}'|^2} \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|}.$$

So

a) $\mathbf{r} - \mathbf{r}' = (5 - 1; 0 - 0) = (4; 0), \quad |\mathbf{r} - \mathbf{r}'| = \sqrt{4^2 + 0^2} = 4\mathrm{cm} = 0.04\mathrm{m}$.

$$\mathbf{E} = 9 \times 10^9 \frac{-12 \times 10^{-9}}{0.04^2} \frac{(0.04; 0)}{0.04} = (-67500; 0) \frac{\mathrm{V}}{\mathrm{m}} = (-67.5; 0) \frac{\mathrm{kV}}{\mathrm{m}}.$$

b) $\mathbf{r} - \mathbf{r}' = (-5 - 1; 0 - 0) = (-6; 0), \quad |\mathbf{r} - \mathbf{r}'| = \sqrt{(-6)^2 + 0^2} = 6\mathrm{cm} = 0.06\mathrm{m}$.

$$\mathbf{E} = 9 \times 10^9 \frac{-12 \times 10^{-9}}{0.06^2} \frac{(0.06; 0)}{0.06} = (-30000; 0) \frac{\mathrm{V}}{\mathrm{m}} = (-30; 0) \frac{\mathrm{kV}}{\mathrm{m}}.$$

c) $\mathbf{r} - \mathbf{r}' = (0 - 1; 5 - 0) = (-1; 5), \quad |\mathbf{r} - \mathbf{r}'| = \sqrt{(-1)^2 + 5^2} = \sqrt{26}\mathrm{cm} = 0.05\mathrm{m}$.

$$\mathbf{E} = 9 \times 10^9 \frac{-12 \times 10^{-9}}{0.05^2} \frac{(-0.01; 0.05)}{0.05} = (8640; -43200) \frac{\mathrm{V}}{\mathrm{m}} = (8.64; -43.2) \frac{\mathrm{kV}}{\mathrm{m}}.$$

Answers: $(-67.5; 0) \frac{\mathrm{kV}}{\mathrm{m}}$, $(-30; 0) \frac{\mathrm{kV}}{\mathrm{m}}$, $(8.64; -43.2) \frac{\mathrm{kV}}{\mathrm{m}}$.

Answer provided by https://www.AssignmentExpert.com