Question

In a Millikan oil-drop experiment (Section 22-8), a uniform electric field of 5.78 x 105 N/C is maintained in the region between two plates separated by 8.56 cm. Find the potential difference (in V) between the plates.

Answer 1

Answer:

The required potential difference is $4.95 \times 10^{4}~V$.

Explanation:

We know, electric field is nothing but the negative gradiant of potential. Mathematically, in three-dimension,

$\vec{E} = -\vec{\nabla}V$

In one dimension, the magnitude of the electric field is

$E = \dfrac{V}{x}$

where '$V$' is the applied voltage and '$x$'is the distance through which the voltage is applied.

Given, $V = 5.78 \times 10^{5}~N~C^{-1}~ and~ x = 8.56~cm = 0.0856~m$.

So the required potential difference is

$V = E \times x = 5.78 \times 10^{5}~N~C^{-1} \times 0.0856~m = 4.95 \times 10^{4}~V$