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Q19.3-26PE

Expert-verifiedFound in: Page 697

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.

- Charge of Sphere – \(1.00{\rm{ }}C\)
- The given potential – \(5.00{\rm{ }}MV\)

**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.**

(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 \(V\)–

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

(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.

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