Chapter 4: Q26E (page 810)

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

Short Answer

Expert verified

(a) If the capacitor is disconnected from the potential source so that the charge on the plates remains constant then the energy will also decrease two times.

(b) If the capacitor remains connected to the potential source so that the potential difference remains the same then the energy will increase two times.

Step by step solution

Relation between capacity, potential and energy

To solve the problem, we need to use the fact that

$C = \frac{ε_{\circ} A}{d}$

Now, the relation between the capacity, potential and energy is given by:

$U = \frac{Q}{2} C = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

Increase and decrease of energy

If the distance is reduced by a factor of two, then the capacitance will increase two time, that is:

$C_{2} = \frac{ε_{\circ} A}{d_{2}} = \frac{e 0 A}{0.5 \times d_{1}} = 2 C_{1}$

Again, if we keep the charges fix and we increase the capacitance the energy will decrease:

$U_{2} = \frac{Q}{2 C_{2}} = \frac{Q}{2 \times 2 C_{1}} = \frac{U_{1}}{2}$

Again, if we keep the voltage constant and change the charge then:

$U_{2} = \frac{1}{2} C_{2} V^{2} = \frac{1}{2} 2 C_{1} V^{2} = 2 U_{1}$

Therefore, the stored energy will increase.

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Short Answer

Step by step solution

Relation between capacity, potential and energy

Increase and decrease of energy

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