Question

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude and direction of the force on a

−7.35 C charge at this location.

Magnitude _____N

Direction?
- north
-south
-east
-west

Answer 1

Approximately
`3.86\; {\rm N}` (given that the magnitude of this charge is
`-7.35\; {\rm \mu C}`.)

Step-by-step explanation:

If a charge of magnitude
`q` is placed in an electric field of magnitude
`E`, the magnitude of the electrostatic force on that charge would be
`F = E\, q`.

The magnitude of this charge is
`q = 7.35\; {\rm \mu C}`. Apply the unit conversion
`1\; {\rm \mu C} = 10^(-6)\; {\rm C}`:

`\begin{aligned} q &= 7.35\; {\mu C} * \frac{10^(-6)\; {\rm C}}{1\; {\mu C}} = 7.35* 10^(-6)\; {\rm C}\end{aligned}`.

An electric field of magnitude
`E = 5.25* 10^(5)\; {\rm N \cdot C^(-1)}` would exert on this charge a force with a magnitude of:

`\begin{aligned}F &= E\, q \\ &= 5.25 * 10^(5)\; {\rm N \cdot C^(-1)} * (-7.35* 10^(-6)\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}`.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.