a charge of 0.05 c moves a negative charge upward due to a 2 n force exerted by an electric field. what is…

a charge of 0.05 c moves a negative charge upward due to a 2 n force exerted by an electric field. what is the magnitude and direction of the electric field?\n0.03 $\frac{n}{c}$ upward\n0.03 $\frac{n}{c}$ downward\n40 $\frac{n}{c}$ upward\n40 $\frac{n}{c}$ downward
Answer
Explanation:
Step1: Recall electric - field formula
The formula for the electric field $\vec{E}=\frac{\vec{F}}{q}$, where $\vec{F}$ is the force on the charge and $q$ is the charge.
Step2: Calculate the magnitude of the electric field
Given $F = 2\ N$ and $q=0.05\ C$. Substitute into the formula: $E=\frac{F}{q}=\frac{2}{0.05}=40\ \frac{N}{C}$.
Step3: Determine the direction of the electric field
The force on a negative charge is in the opposite direction of the electric - field. Since the force on the negative charge is upward, the electric - field is downward.
Answer:
D. $40\ \frac{N}{C}$ downward