The capacitors of capacitance 4 μ F and 6 μ F are connected in series. A potential difference of 500 volt is applied to the outer plates of capacitor system. Then the charge on each capacitor is equal to

Question

The capacitors of capacitance 4 μ F and 6 μ F are connected in series. A potential difference of 500 volt is applied to the outer plates of capacitor system. Then the charge on each capacitor is equal to (A) 600 coulomb (B) 1200 coulomb (C) 1200 micro coulomb (D) 600 micro coulomb.

Tardigrade Admin · Accepted Answer

Total capacitance, Cs = (4 × 6/4 + 6) μ F = 2.4 μF Charge q = VCs = 500 × 2.4 × 106 = 1200 μC

Q. The capacitors of capacitance $4 μ F$ and $6 μ F$ are connected in series. A potential difference of $500 v o lt$ is applied to the outer plates of capacitor system. Then the charge on each capacitor is equal to

600 coulomb

1200 coulomb

1200 micro coulomb

600 micro coulomb.

Total capacitance, $C_{s} = \frac{4 \times 6}{4 + 6} μ$ F = 2.4 $μ$ F
Charge $q = V C_{s} = 500 \times 2.4 \times 1 0^{6}$ = 1200 $μ$ C

Q. The capacitors of capacitance 4μF and 6μF are connected in series. A potential difference of 500volt is applied to the outer plates of capacitor system. Then the charge on each capacitor is equal to