Question:
Two point charges $A$ and $B$, having charges $+ Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of $A$ is transferred to $B$. then force between the charges become
Two point charges $A$ and $B$, having charges $+ Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of $A$ is transferred to $B$. then force between the charges become
Updated On: Jul 1, 2024
- $\frac{16F}{9}$
- $\frac{4F}{3}$
- F
- $\frac{9F}{16}$
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The Correct Option is D
Solution and Explanation
$F =\frac{kQ^{2}}{r^{2}} $
If 25% of charge of A transferred to B then
$ q_{A} =Q - \frac{Q}{4} = \frac{3Q}{4} $ and $ q_{B} = -Q + \frac{Q}{4} = \frac{-3Q}{4} $
$ F_{1} = \frac{kq_{A}q_{B}}{r^{2}} $
$ F_{1} = \frac{k\left(\frac{3Q}{4}\right)^{2}}{r^{2}} $
$ F_{1} = \frac{9}{16} \frac{kQ}{r^{2}} $
$ F_{1} = \frac{9F}{16} $
If 25% of charge of A transferred to B then
$ q_{A} =Q - \frac{Q}{4} = \frac{3Q}{4} $ and $ q_{B} = -Q + \frac{Q}{4} = \frac{-3Q}{4} $
$ F_{1} = \frac{kq_{A}q_{B}}{r^{2}} $
$ F_{1} = \frac{k\left(\frac{3Q}{4}\right)^{2}}{r^{2}} $
$ F_{1} = \frac{9}{16} \frac{kQ}{r^{2}} $
$ F_{1} = \frac{9F}{16} $
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Concepts Used:
Coulomb’s Law
In 1785, french physicist Charles Augustin de Coulomb coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He represented an equation for the force causing the bodies to attract or repel each other which is commonly known as Coulomb’s law or Coulomb’s inverse-square law.
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