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Sample 4: GCE Jun '94 Question 5(d)

A capacitor consisting of two large parallel plates, each of area A and of separation x, has a capacitance given by

C = e_oA/x

The separation x of the plates in such a capacitor is increased by a small amount dx, keeping the charge constant. By considering the corresponding change in the total energy stored in the capacitor, show that the force F exerted by one plate on the other is given by

F = e_oAV²/2x²

where V is the potential difference across the capacitor for a plate separation x. In what direction does the force act?

Thinking Process:

Step 1:

There are 3 equations for energy stored in a capacitor. Which should we use?

The 3 equations are: E = ˝QV, E = ˝CV² and E = ˝Q²/C

Since this is an electrically isolated capacitor and charge is constant, and we know the capacitance formula, we use the third one: E = ˝Q²/C

Step 2:

Initially, the plate separation was x, hence the capacitance C_i=e_oA/x (where C_i stands for initial Capacitance).

After the separation was increased, the plate separation became x+dx. The capacitance became C_f=e_oA/(x +dx) (where C_f stands for final Capacitance).

Step 3:

Using energy E = ˝Q²/C, the initial energy E_i = ˝Q²x/e_oA, and the final energy E_f = ˝Q²(x+dx)/e_oA.

The change in energy is E_f - E_i = ˝Q²dx/e_oA

Step 4:

This increase in energy actually came from work done against electrical forces, which is Fdx.

Equating Fdx = ˝Q²dx/e_oA and canceling the dx on both sides of the equation:

The force F = ˝Q²/e_oA ......... (1)

Step 5:

But we know from the capacitance equation that Q = CV, where C in this case = e_oA/x.

Hence Q =Ve_oA/x

and Q² = V²e_o²A²/x² .......... (2)

Step 6:

Substituting equation (2) into equation (1), we obtain the required equation:

F = e_oAV²/2x²

This force is against the direction of the displacement dx, since the capacitor plates are oppositely charged and attract each other.