Question

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease or stay the same?i) Cii) Qiii) E between the platesiv) delta V

Answer 1

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

$C=\frac{\epsilon_0.A}{d}$

$\therefore C\propto \frac{1}{d}$

In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.

ii)

As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.

iii)

Electric field between the plates is given as:

$E=\frac{\sigma}{\epsilon_0}$

where:

charge density, $\sigma=\frac{Q}{A}$

As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.

iv)

• From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.

• Here we increase the distance so the work done per unit charge increases.