Question

Two capacitors of capacitance 2.0uF and 4.0uF respectively, are each given a charge of 120uC. The positive plates are now connected together, so too are the negative plates. Calculate the new potential difference between the plates of the capacitors.

Answer 1

Explanations & Calculations

• According to the usual equation for capacitors — $\small Q=CV$— generated potential between the plates of $\small 2\mu F$ capacitor is

$\qquad\qquad \begin{aligned} \small V_2&= \small \frac{120\mu C}{2 \mu F}=60\,V \end{aligned}$

• That of the $\small 4\mu F$ capacitor is

$\qquad\qquad \begin{aligned} \small V_4&= \small \frac{120 \mu C}{4 \mu F}=30\,V \end{aligned}$

• For $\bold{\small 2\mu F}$

1) Potential on (+) plate = +30V and on (-) plate= -30V : $\bold{\small \Delta V=60V}$

2) charge on (+) plate = $\bold{\small +60 \mu C}$ and on (-) plate = $\bold{\small -60 \mu F}$

• For $\bold{\small 4\mu F}$

1)+15V and -15V

2) $\bold{\small 60\mu C \,\,\text{and}\,\, 60\mu C}$

• Now the greater potential is with the low valued capacitor.
• When connected (+) to (+) and (-) to (-), what happens is electrons move from the lower (+) plate to the higher (+) plate and from the higher (-) plate to the lower (-) plate until all the like plates (+/+ and -/-) become equipotential plates.
• Equal amounts of electrons move in each connection such that the ultimate potentials are $\bold{\small +V \,\text{and}\,-V}$

• Electron-wise calculation from plate to plate is somewhat difficult and the easy method is to think some $\delta q$ amount of charges move from high potential to low potential (30V $\small \to$ 15V) until potentials get equal.
• Then,

$\qquad\qquad \begin{aligned} \small V_2&= V_4\\ \small \frac{120\mu C-\delta q}{2\mu F}&= \small \frac{120\mu C+\delta q}{4\mu F}\\ \small \delta q&= \small 40\mu C \end{aligned}$

• Now the final potentials are,

$\qquad\qquad \begin{aligned} \small V_{2new}&= \small \frac{(120-40)\mu C}{2\mu F}=\bold{40\,V}\cdots(60V\to40V)\\ \small V_{4new}&= \small \frac{(120+40)\mu C}{4\mu F}=\bold{40\,V}\cdots(30V\to40V) \end{aligned}$