Question

Two parallel-plate capacitors, 6.0 μF each, are connectedin series to a 10 V battery. One of the capacitors is thensqueezed so that its plate separation is halved. Because ofthe squeezing,
(a) how much additional charge is transferred to thecapacitors by the battery and
(b) what is the increase in thetotal charge stored on the capacitors (the charge on thepositive plate of one capacitor plus the charge on the positiveplate of the other capacitor)?

Answer 1

Answer:

a.$60\mu C$

b.$60\mu C$

Explanation:

We are given that

Capacitance of each capacitor=$C=6\mu F$

Potential difference=V=10 V

a.We have to find the additional charge is transferred to the capacitors by the battery.

We know that

$C=\frac{\epsilon_oA}{d}$

Capacitance is inversely proportional to the separation between the plates of capacitor.

When the separation between the plate is halved then the capacitance is doubled.

Therefore, Capacitance, $C'_1$=2 C=$2\times 6=12\mu F$

Initial charge=q=$CV=6\mu F\times 10=60\mu C$

Final charge, q'=$C'_1V=12\times 10=120\mu C$

Additional charge transferred=$q'-q=120-60=60\mu C$

Additional charge transferred=$60\mu C$

b.Initial total charge=$q_i=(C_1+C_2)V=(6\mu+6\mu)(10)=120\mu C$

Final total charge=$q_f=(12+6)(10)=180\mu C$

Increase in total charge=$q_f-q_i=180\mu C-120\mu C=60\mu C$