Question

A proton is moved in a uniform electric field of 1.7 x 10-4 N/c between two point A and B separated by a distance of 0.1m.

(I) What is the potential difference between the two points?

(ii) How much work is done in above process?    ​

Answer 1

Given :

Magnitude of the electric field = 1.7×10⁻⁴ N/C

Distance between the points = 0.1m

Charge of the proton = 1.6×10⁻¹⁹ C

To Find :

a) The potential difference between the two points

b) The work done in the process

Solution :

• Potential difference between two points = E×d

          V = E×d

          V = 1.7×10⁻⁴ × 0.1

          V =  1.7×10⁻⁵ V

The potential difference between the points is  1.7×10⁻⁵ V.

• Work done in the process = q×V

           W = q×V

           W = 1.6×10⁻¹⁹ × 1.7×10⁻⁵

           W = 2.72×10⁻²⁴ J

Work done in the process is 2.72×10⁻²⁴ J.