Question

Locations A and B are in a region of uniform electric field, as shown in Figure 16.67. Along a path from B to A, the change in potential is-2200V. The distance from A to B is 0.28m. What is the magnitude of the electric field in this region?

Short answer

The magnitude of the electric field is$7857.143\mathrm{N}/\mathrm{C}$

Step-by-step solution

Change in the electric potential energy

The magnitude of the electric field for a charged particle (positive or negative) relies on the change in the electric potential energy of that particle.

Also, the potential energy change with the amount of distance the particle travels in the electric filed.

Given data

The change in the potential from B to A is,${\mathrm{\Delta V}}_{\mathrm{BA}}=-2200\hspace{0.33em}\mathrm{V}.$ .

The distance from A to B is, ${\mathrm{\Delta l}}_{\mathrm{AB}}=0.28\hspace{0.33em}\mathrm{m}$.

The magnitude of the electric field

Theformula for the magnitude of the electric field in the regionis given by,

$$\begin{gathered} {\mathrm{\Delta V}}_{\mathrm{BA}}=-\mathrm{E}\times {\mathrm{\Delta I}}_{\mathrm{AB}} \\ \mathrm{E}=\frac{{\mathrm{\Delta V}}_{\mathrm{BA}}}{{\mathrm{\Delta I}}_{\mathrm{AB}}} \\ \mathrm{E}=\frac{-\left(-2200\hspace{0.33em}\mathrm{V}\right)}{0.28\mathrm{m}}\times \frac{1\mathrm{N}/\mathrm{C}}{1\mathrm{V}/\mathrm{m}} \\ \mathrm{E}=7857.143\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the magnitude of the electric field is $7857.143\hspace{0.33em}\mathrm{N}/\mathrm{C}$.