Two parallel plate capacitors A and B having capacitance 1µF and 5 µF are charged separately to the same potential 100V. They are then connected such that +ve plate of A is connected to –ve plate of B. Find the charge on each capacitor and total loss of energy in the capacitors

Dear Student,

Please find below the solution to the asked query:

Given information,

$$\begin{gathered} C_A=1 \mu F, C_B=5 \mu F, V_A=V_B=100 V, \\ So, \\ {q}_A=C_A{V}_A=1\times {10}^{-6}\times 100={10}^{-4} C \\ {q}_B=C_B{V}_B=5\times {10}^{-6}\times 100=5\times {10}^{-4} C \end{gathered}$$

After connecting,

$$\begin{gathered} Total charge (on B_1 and A_1) Q=-5\times {10}^{-4}+1\times {10}^{-4}=-4\times {10}^{-4} C \\ Total charge (on B_2 and A_2) Q=5\times {10}^{-4}-1\times {10}^{-4}=+4\times {10}^{-4} C \end{gathered}$$

Hence after connecting both the plates A₁ and B₁ are having negative charge and the plates A₂ and B₂ are having positive charges. So, the combination of A & B are in parallel.

Therefore,
$V=\frac{Q}{C}=\frac{4\times {10}^{-4}}{6\times {10}^{-6}}=\frac{400}{6} Volt$
Hence charges on A & B after connection

$$\begin{gathered} q_A\text{'}=C_{A}V=1\times {10}^{-6}\times \frac{400}{6}=\frac{2}{3}\times {10}^{-4} C \\ {q}_B\text{'}=C_{B}V=5\times {10}^{-6}\times \frac{400}{6}=\frac{1}{3}\times {10}^{-3} C \end{gathered}$$

​Final Energy is,

$$\begin{gathered} U_F=\frac{1}{2}C{V}^2=\frac{1}{2}\times 6\times {10}^{-6}\times {\left(\frac{400}{6}\right)}^2 \\ \Rightarrow U_F=\frac{4}{3}\times {10}^{-2} J \end{gathered}$$

Initial Energy is,

$$\begin{gathered} U_i=\frac{1}{2}{C}_A{V}_A^2+\frac{1}{2}{C}_B{V}_B^2=\frac{1}{2}\times 1\times {10}^{-6}\times {\left(100\right)}^2+\frac{1}{2}\times 5\times {10}^{-6}\times {\left(100\right)}^2 \\ \Rightarrow U_F=3\times {10}^{-2} J \end{gathered}$$

Therefore, the loss of energy is,

$$\begin{gathered} ∆U=U_F-U_i=\left(3-\frac{4}{3}\right)\times {10}^{-2}=\frac{5}{3}\times {10}^{-2} J \\ \Rightarrow ∆U=1.67\times {10}^{-2} J \end{gathered}$$
 

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