Question

In a Millikan oil-drop experiment, a uniform electric field of 5.71 x 10^5 N/C is maintained in the region between two plates separated by 6.49 cm. Find the potential difference (in V) between the plates.

Answer 1

Answer:

V = 3.71×10⁴ V

Explanation:

Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.

The S.I unit of potential difference is Volt (V).

The expression for potential difference is

V = E×d.............................. Equation 1

Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates

Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.

Substitute into equation 1

V = 5.71×10⁵×0.0649

V = 3.71×10⁴ V

Answer 2

Answer:

37057.9V

Explanation:

Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.

Electric potential (E) = Potential Difference (V)/distance between plates(d)

Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m

Since E = V/d

V = Ed

V = 5.71×10^5×0.0649

V = 37057.9Volts

The potential difference (in V) between the plates is 37057.9V