Question

An electron is released in a uniform electric field, and it experiences an electric force of 2.36 x10--13 N directed toward the north. Find the magnitude and direction of the electric field.

Answer 1

Answer:

The magnitude of the electric field is 1475000 N/C and the  direction of the electric field is toward north.

Explanation:

Given that,

Electric force $F=2.36\times10^{-13}\ N$

The direction of force is toward the north.

We need to calculate the magnitude of the electric field

Using formula of electric field

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

Where, f = electric force

q = charge

Put the value into the formula

$E=\dfrac{2.36\times10^{-13}}{1.6\times10^{-19}}$

$E=1475000\ N/C$

The direction of the electric field is toward the direction of the force so,

The direction of the electric field is toward north.

Hence, The magnitude of the electric field is 1475000 N/C and the  direction of the electric field is toward north.