Question

In a Millikan oil-drop experiment (Module 22-6), a uniform electric field of $\mathbf{1}\mathbf{.92}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{\text{\hspace{0.17em}N}}\mathbf{/}\mathbf{\text{C}}$ is maintained in the region between two plates separated by $\mathbf{1.50}\mathbf{\text{\hspace{0.17em}cm}}$ . Find the potential difference between the plates.

Short answer

The potential difference between the plates is, $2.9\text{\hspace{0.17em}kV}$.

Step-by-step solution

Step 1: Identification of the given data

The uniform electric field is,$\mathrm{E}=1.92\times {10}^5\text{\hspace{0.17em}N}/\text{C}$.

The separation between the two plates is,$\mathrm{R}=0.015\text{\hspace{0.17em}m}$

Understanding the concept

According to the definition, the electric potential gradient ‘$\mathbf{E}$’ is given as:

role="math" localid="1662732704306" $\mathbf{E}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{\nu }}{\mathbf{\partial }\mathbf{R}}$

Or

$\begin{array}{c}\mathbf{}\int \partial V\mathbf{=}\mathbf{E}\mathbf{\times }\int \partial R\\ \mathbf{V}\mathbf{=}\mathbf{ER}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Step 3: Calculate the potential difference between the plates

The potential difference between the plates is expressed is,

$\mathrm{V}=\mathrm{ER}$

Substitute all the value in the above equation.

Hence the potential difference between the plates is,$2.9\text{\hspace{0.17em}kV}$ .