Question

A proton is released in a uniform electric field, and it experiences an electric force of 2.07 x 10^-14 N toward the south. Part A) What is the magnitude of the electric field? Part B) What is the direction of the electric field? O west O east O south O north

Answer 1

Answer:

The magnitude of the electric field is 129375 N/C toward south.

Explanation:

Given that,

Electric force $F=2.07\times10^{-14}\ N$

(A). We need to calculate the magnitude of the electric field

Using formula of electric field

$F = qE$

$E=\dfrac{F}{q}$

Where, q = charge of proton

E = electric field

$E=\dfrac{2.07\times10^{-14}}{1.6\times10^{-19}}$

$E=129375\ N/C$

(B). The direction of the electric field is toward the direction of the force.

So, The direction of the electric field is toward south

Hence, The magnitude of the electric field is 129375 N/C toward south.