Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 28 - The Electric Potential - Exercises and Problems - Page 837: 63

Answer

$2.13\;\rm kV$

Work Step by Step

We know that the potential at any point is the superposition of the potentials due to all charges [as the author mentioned as a hint]. Let's call the big charged sphere as 1 and the other one as 2. So, the potential at point a is given by $$V_a=V_1+V_2$$ From the given figure, $$V_a=\dfrac{kQ_1}{R_1}+\dfrac{kQ_2}{1-R_1}$$ $$V_a=k\left[\dfrac{ Q_1}{R_1}+\dfrac{ Q_2}{1-R_1}\right]\tag 1$$ The potential at point b is given by $$V_b=V_1+V_2$$ From the given figure, $$V_b=\dfrac{kQ_1}{1-R_2}+\dfrac{kQ_2}{R_2}$$ $$V_b=k\left[\dfrac{ Q_1}{1-R_2}+\dfrac{Q_2}{R_2}\right]\tag 2$$ Thus, the potential difference between points a and b is given by $$\Delta V=V_b-V_a$$ Plug from (1) and (2), $$\Delta V=k\left[\dfrac{ Q_1}{1-R_2}+\dfrac{Q_2}{R_2}\right]-k\left[\dfrac{ Q_1}{R_1}+\dfrac{ Q_2}{1-R_1}\right]$$ Plugging the known; $$\Delta V=(9\times 10^9)\left[\dfrac{ 100\times 10^{-9}}{1-0.05}+\dfrac{25\times 10^{-9}}{0.05}\right]-\\ (9\times 10^9)\left[\dfrac{ 100\times 10^{-9}}{0.30}+\dfrac{ 25\times 10^{-9}}{1-0.30}\right]$$ $$\Delta V=\color{red}{\bf 2126}\;\rm V$$ It is obvious now that $V_b\gt V_a$ since the final result is positive.
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