Question

A 7.0 µF capacitor and a 5.0 µF capacitor are connected in series across a 5.0 kV potential difference. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor (in mC) and the voltage across each capacitor (in V). (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

Answer 1

The charge on the first capacitor is 14.59 mC

The voltage across the first capacitor is 2084 V

The charge on the second capacitor is 14.59 mC

The voltage across the second capacitor is 2918 V

Step-by-step explanation:

Given;

first capacitor, C₁ = 7.0 µF

second capacitor, C₂ = 5.0 µF

potential difference, V = 5000 V

The equivalent capacitance is given by;

`(1)/(C_(eq))= (1)/(C_1) +(1)/(C_2)\\\\(1)/(C_(eq))= (C_1 + C_2)/(C_1C_2)\\\\C_(eq) = (C_1C_2)/(C_1 +C_2)\\\\ C_(eq) = ((7*10^(-6))(5*10^(-6)))/((7*10^(-6) \ +\ 5*10^(-6)))\\\\C_(eq) = 2.917*10^(-6) \ F`

The charge on the first capacitor is given by;

`Q_1 = C_(eq) V\\\\Q_1 = 2.917*10^(-6) *5000\\\\Q_1 = 0.01459 \ C\\\\Q_1 = 14.59 \ mC`

The voltage across the first capacitor is given by;

`V_1 = (Q_1)/(C_1) \\\\V_1 = (0.01459)/(7*10^(-6))\\\\ V_1 = 2084.286 \ V`

V₁ = 2084 V

The charge on the second capacitor is given by;

`Q_2 = C_(eq) V\\\\Q_2 = 2.917*10^(-6) *5000\\\\Q_2 = 0.01459 \ C\\\\Q_2 = 14.59 \ mC`

The voltage across the second capacitor is given by;

`V_2 = (Q_2)/(C_2)\\\\ V_2 = (0.01459)/(5*10^(-6))\\\\V_2 = 2918 \ V`