Question

After the switch in the circuit below has been closed for a long time, what will the charge be on the $4.00$ microfarad capacitor and on the $6.00$ microfarad capacitor?

• A. Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $100.0microcoulombs-100.0microcoulombs$
• B. Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $60.0microcoulombs-40.0microcoulombs$
• C. Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $40.0microcoulombs-60.0microcoulombs$
• D. Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $8.0microcoulombs-6.0microcoulombs$
• E. Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $6.0microcoulombs-8.0microcoulombs$

Answer 1

The correct option is C Charge on $4.00$ microfarad Capacitor - Charge on $6.00$ microfarad Capacitor, $40.0microcoulombs-60.0microcoulombs$

After the switch has been closed for a long time, the capacitors get gully charged and hence no current flows through the capacitors.

$⟹$ The entire voltage of the battery is developed across both capacitors.

$\therefore$ Charge on $4.00$ $\mu F$ capacitor $Q_1=C_{1}V=4.00\times 10.0=40.0$ $\mu$C

Similarly charge on $6.00$ $\mu F$ capacitor $Q_2=C_{2}V=6.00\times 10.0=60.0$ $\mu$C