Question

A capacitor of capacitance of 5 micro faraday is charged to potential of 500 volts .Then it is disconnected from the battery and connected to uncharged capacitor of capacitance 3 microfaraday . calculate the common potential , charge on each capacitor and the loss of energy?

Answer 1

Dear student
$$\begin{gathered} Formulaforcommonpotentialis \\ {V}_c=\frac{sumofchargesonbothplatesofthetwocapacitorsconnectedtogetheralongwithsignofcharge}{sumofcapci\tan ces} \\ Chargeonthecapacitorwhichischargedbythecellis{q}_1=5\mu F\times \left(500\right) \\ chargeontheothercapacitoris{q}_2=0 \\ {V}_c=\frac{q_1+q_2}{C_1+C_2}=\frac{5\mu F\times \left(500\right)}{(5+3)\mu F}=312.5Volt \\ chargeon5\mu Fis \\ {q}_1=5\mu F\times 312.5=1562.5\mu C \\ chargeon3\mu Fis \\ {q}_2=3\mu F\times 312.5=937.5\mu C \\ b)heatproduced=\frac{1C_1\times C_2}{2(C_1+C_2)}(V_1-V_2{)}^2=\frac{1\left(5\right)\times \left(3\right)}{2\times 8}\times (500-0{)}^2\times {10}^{-6}J=\frac{15}{16}\times 25\times {10}^4\times {10}^{-6}=\frac{15}{64}=0.234J \\ Regards \end{gathered}$$