Answer to Question #159171 in Electric Circuits for Tony

Question

The capacitance of a certain variable capacitor may be varied between limits of 1 x 10^(-10) F and 5 x 10^(-10) F by turning a knob attached to the movable plates. The capacitor is set to 5 x 10^(-10) F, and is charged by connecting it to a battery of EMF 200V.
(i) What is the charge on the plates?The battery is then disconnected and the capacitance changed to 1 x 10^(-10) F.
(ii) Assuming that no charge is lost from the plates, what is now the PD between them?
(iii) How much mechanical work is done against electrical forces in changing the capacitance?

Answer 1

Explanations & Calculations

• A Capacitor, when connected to a voltage source, stores electrical charges on both its conducting plates until the potential difference between those plates get equal to that of the external source.
• For a given capacitance & a voltage source, it stores a fixed amount of charges according to the equation $\small Q=CV$
• Once the voltage source is removed, the capacitor retains those stored charges on it & the potential difference between the plates still remains in the previous value.
• According to the definition of the capacitance—how many charges stores under a unit voltage—what happens after capacitance decreased is that those stored charges equivalents to some higher potential difference or that much of charges could be stored in that capacitor under comparatively higher potential difference.
• Energy stored in a capacitor is given by $\small W=\large\frac{1}{2}\frac{Q^2}{C}$. According to that energy stored in it increases as the capacitance is decreased.
• The assumption made here is "no charge loss".

1)

• Charges stored in the capacitor in the first arrangement,

$\qquad\qquad \begin{aligned} \small Q_i&= C_iV_i\\ &= \small (5\times10^{-10}F)(200V)\\ &=\small 1\times10^{-7}\,C \end{aligned}$

2)

• After the capacitance decreased,new potential difference would be

$\qquad\qquad \begin{aligned} \small Q_i&= \small C_{new}V_{new}\\ \small 1\times10^{-7}&= \small (1\times10^{-10}F).V_{new}\\ \small V_{new}&= \small 10^3\\ \small V_{new}&=\small \bold{1000V} \end{aligned}$

3)

• Work needs to be done to change the capacitance is

$\qquad\qquad \begin{aligned} \small \Delta W&= \small W_{new}-W_i\\ &= \small \frac{1}{2}\frac{Q^2}{C_{new}}-\frac{1}{2}\frac{Q^2}{C_i}\\ &= \small \frac{Q^2}{2}\bigg[\frac{1}{C_{new}}-\frac{1}{C_i}\bigg] \\ &= \small \frac{(1\times10^{-7}C)^2}{2}\bigg[\frac{1}{1\times10^{-10}F}-\frac{1}{5\times10^{-10}F}\bigg]\\ &=\small\bold{4\times10^{-5}J} \end{aligned}$