a charge of +0.08 c moves to the right due to a 4 n force exerted by an electric field. what is the…

a charge of +0.08 c moves to the right due to a 4 n force exerted by an electric field. what is the magnitude and direction of the electric field?\n0.32 n/c left\n0.32 n/c right\n50 n/c left\n50 n/c right

a charge of +0.08 c moves to the right due to a 4 n force exerted by an electric field. what is the magnitude and direction of the electric field?\n0.32 n/c left\n0.32 n/c right\n50 n/c left\n50 n/c right

Answer

Explanation:

Step1: Recall electric - field formula

The formula for the electric field is $E=\frac{F}{q}$, where $E$ is the electric field, $F$ is the force on the charge, and $q$ is the charge.

Step2: Substitute values

Given $F = 4\ N$ and $q=0.08\ C$. Substitute these values into the formula: $E=\frac{4}{0.08}=50\ N/C$.

Step3: Determine direction

The force on a positive charge is in the direction of the electric - field. Since the positive charge moves to the right due to the force, the electric - field is also to the right.

Answer:

D. 50 N/C right