A neutral conducting solid sphere of radius $R$ has two spherical cavities of radii $a$ and $b$ as shown in the figure. The center-to-center distance between the two cavities is $c$. $q_a$ and $q_b$ charges are placed at the centers of cavities respectively. The force between $q_a$ and $q_b$ is
Given: A neutral conducting solid sphere with two spherical cavities. The radii of the cavities are \( a \) and \( b \), and the center-to-center distance between the two cavities is \( c \). Charges \( q_a \) and \( q_b \) are placed at the centers of the cavities respectively.
To find: The force between the charges \( q_a \) and \( q_b \).
Key Concept: In a conductor, charges redistribute themselves on the outer surface in such a way that the electric field inside the conductor is zero. Furthermore, the presence of cavities in the conductor does not affect the electric field within the conducting material itself. The force between the charges in the cavities is influenced by the conducting nature of the sphere and its symmetry.
Solution: Since the conductor is neutral, the electric field inside the conductor is zero, and the charges on the cavities do not directly interact with each other in terms of electrostatic force. The conducting sphere ensures that the electric fields created by \( q_a \) and \( q_b \) do not interact in the usual way. Thus, the force between the charges \( q_a \) and \( q_b \) is: \[ \text{Force} = 0. \]
Final Answer: The force between the charges \( q_a \) and \( q_b \) is zero.
