Electric Charges And Fields

Let's assume the original charges are q₁ and q₂, and the distance between them is l. The initial force is given by:
F_initial = k *
Now, if one of the charges is doubled (let's say q₁) and the distance between them is halved ( ), the new force is given by:
F_new = k *
F_new = 4 * k *
F_new = 4 * k *
F_new = 16 * k *
Comparing the new force to the initial force:
=
= 16
Therefore, the magnitude of the force becomes 16 times the initial force.
The answer is (A) 16.

The flux linked with a closed surface is given by Gauss's Law.
Φ =
In the given scenario, there is a charge q within the hollow cylinder. Let's consider the curved surface B and the plane surface A separately.
For the curved surface B, the electric flux associated with it is ϕ.
For the plane surface A, the flux linked with it can be calculated as follows:
Φ(A) = Φ - Φ(B)
Substituting the values, we get:
Φ(A) = - ϕ
Therefore, the correct answer is (C) .