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102P

Expert-verifiedFound in: Page 716

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A charge of **${\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{8}}{\mathbf{}}{\mathbf{C}}$** lies on an isolated metal sphere of radius 16.0 cm****. With** ${\mathbf{V}}{\mathbf{=}}{\mathbf{0}}$** at infinity, what is the electric potential at points on the sphere’s surface?**

The electric potential at points on the sphere’s surface is $\mathrm{V}=844\mathrm{V}$.

For the isolated metal sphere:

The radius, $\mathrm{R}=16.0\mathrm{cm}$

The value of the charge $\mathrm{q}=1.50\times {10}^{-8}\mathrm{C}$

At infinity, $\mathrm{V}=0$

**Use the equation of the electric potential at points on the sphere’s surface.**

**The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field.**

Formula:

The electric potential is define by,

$\begin{array}{rcl}\mathrm{V}& =& \frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{\mathrm{q}}{\mathrm{R}}\\ & =& \mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Here, V is electric potential, R is distance between the point charges, q is charge, ${\mathrm{\epsilon}}_{0}$ is the permittivity of free space, and k is the Coulomb’s constant having a value $9\times {10}^{9}\mathrm{N}.{\mathrm{m}}^{2}/{\mathrm{C}}^{2}$.

The electric potential on the sphere is,

$\begin{array}{rcl}\mathrm{V}& =& \frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{\mathrm{q}}{\mathrm{R}}\\ & =& \mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{V}& =& 9.0\times {10}^{9}\times \frac{1.5\times {10}^{-8}}{0.16}\\ & =& 843.7\mathrm{volts}\\ & \approx & 844\mathrm{volts}\end{array}$

Hence, the electric potential at points on the sphere’s surface is 844 V.

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