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Q77P

Expert-verifiedFound in: Page 715

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 ${\mathbf{1}}{\mathbf{.92}}{\mathbf{}}{\mathbf{\times}}{\mathbf{}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{\text{\hspace{0.17em}N}}}{\mathbf{/}}{\mathbf{\text{C}}}$ 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.**

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

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, $\mathrm{R}=0.015\text{\hspace{0.17em}m}$

**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**

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

The potential difference between the plates is expressed is,

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

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{V}=1.92\times {10}^{5}\text{\hspace{0.17em}N}/\text{C}\times 0.015\text{\hspace{0.17em}m}\\ =2.88\times {10}^{3}\text{\hspace{0.17em}V}\\ =2.9\text{\hspace{0.17em}kV}\end{array}$

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

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