Question

In the Millikan oil-drop experiment (see Section 22-8), a uniform electric field of 1.92 ✕ 105 N/C is maintained in the region between two plates separated by 1.50 cm. Find the potential difference between the plates.

Answer 1

Answer:

2.88×10³ V.

Explanation:

Potential difference: This can be defined as the work done in bringing a unit charge from infinity to any point in an electric field. The S.I unit of electric potential is V.

The formula of electric potential is given as,

V = E×r.................................................. Equation 1

Where V = Electric potential, E = Electric Field, r = distance between the plates.

Given: E = 1.92×10⁵ N/C, r = 1.5 cm = 0.015 m.

Substitute into equation 1

V = 1.92×10⁵×0.015

V = 2.88×10³ V.

Hence the potential difference = 2.88×10³ V.

Answer 2

Answer:

2.88 x 10³V or 2880V

Explanation:

The electric field (E) between the plates is related to the potential difference (V) and the distance (d) between the plates as follows;

V = E x d       -------------------(i)

According to the question;

E = 1.92 x 10⁵ N/C

d = 1.50cm = 0.015m

Substitute these values into equation (i) as follows;

V = 1.92 x 10⁵ x 0.015

V = 0.0288 x 10⁵ V

V = 2.88 x 10³V or 2880V

Therefore, the potential difference between the plates is 2.88 x 10³V or 2880V