**5) A 5 \muF capacitor is charged to a potential difference of 110 V and then connected in parallel with an uncharged 3 \muF capacitor. Calculate the potential difference across the parallel capacitor.**

Q_1 = C_1 \times V_1 = 5 \times 10^{-6} \times 110 = 5.5 \times 10^{-4} C

Both the capacitors are connected in parallel. Therefore voltage across both the capacitor will remain same.

V_1 = V_2

\dfrac{q_1}{C_1} = \dfrac{q_2}{C_2}

Now capacitor C_2 is uncharged, therefore the charge will flow from C_1 to C_2 and this flow of charge will stop when the potential difference across C_1 and C_2 will become equal i.e. V_1 = V_2. Capacitor C_1 will loose charge say q_2 and capacitor C_2 will gain that charge q_2. Hence new charge on C_1 will become q_1.

\therefore Charge on C_1 will be q_1 = Q_1 – q_2

\dfrac{Q_1 – q_2}{C_1} = \dfrac{q_2}{C_2}

\dfrac{(5.5 \times 10^{-4}) – q_2}{5 \times 10^-6} = \dfrac{q_2}{3 \times 10^-6}

q_2 = 2.06525 \times 10^{-4}

V_1 = V_2 = \dfrac{q_2}{C_2} = 68.75 V

**Method 2**

Q_1 = C_1 \times V_1 = 5.5 \times 10^{-4} C

When the capacitors are connected in parallel, the total capacitance is 8 \mu F and the charge of 0.01 C is divided between the two capacitor.

Q = CV

V = \dfrac{Q}{C}

V = \dfrac{5.5 \times 10 ^{-4}}{8 \times 10^{-6}}

V = 68.75 V