a +0.05 c charge is placed in a uniform electric field pointing downward with a strength of 100…

a +0.05 c charge is placed in a uniform electric field pointing downward with a strength of 100 $\frac{n}{c}$. determine the magnitude and direction of the force on the charge.\n5 n downward\n5 n upward\n2000 n downward\n2000 n upward

a +0.05 c charge is placed in a uniform electric field pointing downward with a strength of 100 $\frac{n}{c}$. determine the magnitude and direction of the force on the charge.\n5 n downward\n5 n upward\n2000 n downward\n2000 n upward

Answer

Explanation:

Step1: Recall the formula for electric force

The formula for the force on a charge in an electric field is $F = qE$, where $q$ is the charge and $E$ is the electric - field strength.

Step2: Substitute the given values

Given $q = 0.05\ C$ and $E=100\ \frac{N}{C}$. Then $F=(0.05\ C)\times(100\ \frac{N}{C}) = 5\ N$.

Step3: Determine the direction of the force

Since the charge is positive ($q>0$) and the electric field points downward, the force on a positive charge in an electric field has the same direction as the electric field. So the force points downward.

Answer:

5 N downward