Question

An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the magnitude and the direction of the electric field?

Answer 1

Step-by-step explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .