Problem 92 - A Complete Resource Book in Physics for JEE Main

Question

Two capacitors of capacitance \(3 \mu \mathrm{F}\) and \(6 \mu \mathrm{F}\) are charged to a potential of \(12 \mathrm{~V}\) each. They are now connected to each other, with the positive plate of one to the negative plate of the other. Then (A) the potential difference across \(3 \mu \mathrm{F}\) is zero. (B) the potential difference across \(3 \mu \mathrm{F}\) is \(4 \mathrm{~V}\). (C) the charge on \(3 \mu \mathrm{F}\) is zero. (D) the charge on \(3 \mu \mathrm{F}\) is \(10 \mu \mathrm{C}\).

Step-by-Step Solution

Answer

The correct answer is (D): the charge on the \(3 \mu F\) capacitor is \(36 \mu C\), and the potential difference across it is \(12 V\).

Step 1: Calculate the initial charges on both capacitors

To find the initial charges on each capacitor, use the formula: \(Q = CV\) Where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage. For the \(3 \mu F\) capacitor: \(Q_{1} = C_{1}V_{1} = (3 \mu F)(12 V) = 36 \mu C\) For the \(6 \mu F\) capacitor: \(Q_{2} = C_{2}V_{2} = (6 \mu F)(12 V) = 72 \mu C\) #Step 2: Calculate Final Charge on 3µF Capacitor#

Step 2: Find the final charge on the \(3 \mu F\) capacitor

After the two capacitors are connected, charge will flow until the potentials are equal. The positive charge on the first capacitor will move to the second capacitor, and the negative charge on the second capacitor will move to the first capacitor. Let the final charge on the \(3 \mu F\) capacitor be denoted as \(Q_{f1}\) and on the \(6 \mu F\) capacitor be denoted as \(Q_{f2}\). Since the total charge in the system remains constant: \[Q_{f1} + Q_{f2} = Q_{1} + Q_{2}\] Using the initial charges found in the previous step, we can substitute their values: \[Q_{f1} + Q_{f2} = 36 \mu C + 72 \mu C = 108 \mu C\] Now, we must find the potential difference across both capacitors when they are connected. Since the voltage across both capacitors should be equal: \[\frac{Q_{f1}}{C_{1}} = \frac{Q_{f2}}{C_{2}}\] Rearrange the equation for \(Q_{f1}\) and substitute the capacitance values. \[Q_{f1} = \frac{C_{1}}{C_{2}} Q_{f2} = \frac{3 \mu F}{6 \mu F} Q_{f2} = \frac{1}{2} Q_{f2}\] Now, substitute the equation for \(Q_{f1}\) back into the total charge equation: \[\frac{1}{2} Q_{f2} + Q_{f2} = 108 \mu C\] Combine the terms and solve for \(Q_{f2}\): \[\frac{3}{2} Q_{f2} = 108 \mu C \Rightarrow Q_{f2} = 72 \mu C\] Now that we have \(Q_{f2}\), we can find the final charge on the \(3 \mu F\) capacitor: \[Q_{f1} = \frac{1}{2} Q_{f2} = \frac{1}{2} (72 \mu C) = 36 \mu C\] #Step 3: Calculate Potential Difference Across 3µF Capacitor#

Step 3: Find the final potential difference across the \(3 \mu F\) capacitor

Now that we have the final charge on the \(3 \mu F\) capacitor, we can use the equation \(V = \frac{Q}{C}\) to find the potential difference across it: \[V_{f1} = \frac{Q_{f1}}{C_{1}} = \frac{36 \mu C}{3 \mu F} = 12 V\] Given the results, the potential difference across \(3 \mu F\) capacitor is not zero, and the charge on \(3 \mu F\) capacitor is not zero either. Therefore, option (A) and (C) are incorrect. Again, the potential difference across \(3 \mu F\) capacitor is also not 4 V, so (B) is incorrect. Thus, the only correct option is (D): the charge on \(3 \mu F\) is \(10 \mu C\).