Two identical conducting spheres $A$ and $B$, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is $F$. A third identical conducting sphere, $C$, is uncharged. Sphere $C$ is first touched to $A$, then to $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to :
- $F$
- $\frac{3F}{4}$
- $\frac{3F}{8}$
- $\frac{F}{2}$
The Correct Option is C
Solution and Explanation
Therefore, force between spheres $A$ and $B$ is
$F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{d^{2}}$...(1)
When spheres $A$ and $C$ are touched and then separated, charge on each will be $\frac{q+0}{2}$, i.e. $\frac{q}{2} .$
Now sphere $B$ is touched with sphere $C$, charge on each will be $\left[\frac{q+\frac{q}{2}}{2}\right]=\frac{3 q}{4}$.
Now force between sphere $A$ and sphere $B$ will be
$F'=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\frac{q}{2} \cdot \frac{3 q}{4}}{d^{2}}=\frac{3}{8} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{d^{2}}$
$\Rightarrow F'=\frac{3}{8} F$
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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.
As per Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges regarded to be point charges.
Coulomb’s Law has an abundant application to modern life, from Xerox machines to laser printers, to powder coating.