Question

A capacitor is connected to a battery. The force of attraction between the plates when the separation between them is halved:
(A) remains the same
(B) becomes eight times
(C) becomes four times
(D) becomes two times

Answer 1

Hint: When the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates. We can use the boundary conditions to solve this problem.

Complete step by step solution:
We know capacitance of a parallel plate capacitor is given by \[C=\dfrac{{{\varepsilon }_{0}}A}{d}\], and d here is the separation between the two plates. So, when the separation is halved the capacitance becomes twice. Mathematically, we can write that,
\[
 C=\dfrac{{{\varepsilon }_{0}}A}{d'} \\
 \Rightarrow C=\dfrac{{{\varepsilon }_{0}}A}{\dfrac{d}{2}} \\
 \Rightarrow C=\dfrac{2{{\varepsilon }_{0}}A}{d} \\
\]
Also, Q=CV
Since capacitance has become two times, the charge stored also becomes two times. The relationship between the electric field and potential difference between the two plates is given by \[E=\dfrac{V}{d}\].
Since the separation has become half, the value of the electric field becomes two times. Now from Coulomb's law
\[F=qE\]
Now charge has become twice and electric field has also, become two times, so,
\[
 F=q'E' \\
 \Rightarrow F=2q\times 2E \\
 \Rightarrow F=4qE \\
\]
Since charge has become two times and the electric field has also become two times, so, the force by this relation becomes 4 times.

So, the correct option is (C)

Note:
In this problem the space between the two plates of the capacitor was empty and we have assumed vacuum is between the two plates. But if space was filled with some dielectric then we will have to take into account the role of dielectric too.