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Q19.3-26PE

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Found in: Page 697

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) A sphere has a surface uniformly charged with $$1.00{\rm{ }}C$$. At what distance from its centre is the potential $$5.00{\rm{ }}MV$$? (b) What does your answer imply about the practical aspect of isolating such a large charge?

(a) The distance at which a sphere having uniformly charged surface has potential $$5.00{\rm{ }}MV$$, is $$r = 1.80\;km$$.

(b) The practical aspect represents that $$1$$ coulomb is also a big charge as a lightning bolt only carries $$15$$ coulombs of charge.

See the step by step solution

## Step 1: Given Information

• Charge of Sphere – $$1.00{\rm{ }}C$$
• The given potential – $$5.00{\rm{ }}MV$$

## Step 2: Concept Introduction

Electric Potential Due to a Single Charge: Place a test charge $$q$$ at the desired site, and then calculate the electric potential energy $${U_{Qq}}$$ of the system containing the test charge and the source charge that generates the field to determine the electric potential owing to a single charge at that location.

The ratio - $$V = \frac{{{U_{Qq}}}}{q} = \frac{{kQ}}{r}...(1)$$ determines the electric potential there, where $$k = 8.99 \times {10^9}{\rm{ }}N \times {m^2}/C$$ is a proportionality constant known as the Coulomb's constant. The joule/coulomb $$\left( {J/C} \right)$$, often known as the volt $$\left( V \right)$$ is the unit of electric potential.

## Step 3: Calculation for the distance

(a)

The electric potential a distance $$r$$ from the proton is found from Equation $$(1)$$ –

$$V = \frac{{kq}}{r}$$

Solving for $$r$$, it is obtained –

$$r = \frac{{kq}}{V}$$

Entering the values for $$k,q$$, and $$V$$–

$$\begin{array}{c}r = \frac{{\left( {8.99 \times {{10}^9}\;N \times {m^2}/C} \right)(1.00C)}}{{5.00 \times {{10}^6}\;V}}\\ = 1.80 \times {10^3}\;m\\ = \left( {1.80 \times {{10}^3}\;m} \right)\left( {\frac{{1\;km}}{{1000\;m}}} \right)\\ = 1.80\;km\end{array}$$

## Step 4: Practical aspect of isolating large charge

(b)

A charge of $$1$$ coulomb on the sphere is extremely large amount of charge. To get a feeling of how big a charge of $$1$$ coulomb is, you should know that a typical lightning bolt transfers only $$15$$ coulombs of charge. In this problem, the charge generates a potential of $$5.00{\rm{ }}MV$$ a distance $$1.80\;km$$ away from it which is a huge distance.

Therefore, the distance is obtained as $$r = 1.80\;km$$and it represents that the charge is very large when implying practically.