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Expert-verified Found in: Page 715 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In a Millikan oil-drop experiment (Module 22-6), a uniform electric field of ${\mathbf{1}}{\mathbf{.92}}{\mathbf{}}{\mathbf{×}}{\mathbf{}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{\text{\hspace{0.17em}N}}}{\mathbf{/}}{\mathbf{\text{C}}}$ 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.

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

See the step by step solution

## Step 1: Identification of the given data

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

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

## Step 2: 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

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

## 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, $2.9\text{\hspace{0.17em}kV}$ . ### Want to see more solutions like these? 