Problem 175 - A Complete Resource Book in Physics for JEE Main

Question

Two capacitors \(C_{1}\) and \(C_{2}\) are charged to \(120 \mathrm{~V}\) and \(200 \mathrm{~V}\), respectively. It is found that by connecting them together, the potential on each one can be made zero. Then (A) \(3 C_{1}=5 C_{2}\) (B) \(3 C_{1}+5 C_{2}=0\) (C) \(9 C_{1}=4 C_{2}\) (D) \(5 C_{1}=3 C_{2}\)

Step-by-Step Solution

Answer

The correct relationship between the capacitors is \(3C_{1} = -5C_{2}\). However, this is not one of the options, thus it could be an error in the problem statement or the student's understanding of potential could be wrong. The potential being zero does not mean that the charges are zero but rather that the voltages are equal.

Step 1: Identifying Capacitors' Charges

Identify the charge \(Q_{1}\) on \(C_{1}\) and \(Q_{2}\) on \(C_{2}\) using the formula \(Q = CV\). For \(C_{1}\), \(Q_{1} = 120C_{1}\) and for \(C_{2}\), \(Q_{2} = 200C_{2}\). So, the charges on the capacitors before they are connected together are \(Q_{1}\) and \(Q_{2}\).

Step 2: Applying Charge Conservation

When the capacitors are connected together, the total charge of the system has to be conserved. Therefore, you take the sum of charges \(Q_{1}\) and \(Q_{2}\) which should be equal to 0 in our case, keeping in mind that potential on each one is made zero. Therefore, \(120C_{1} + 200C_{2} = 0\).

Step 3: Solving the Equation

We solve this equation to get the relationship between \(C_{1}\) and \(C_{2}\). Dividing the whole equation by \(20C_{2}\) we get \(6C_{1}+10C_{2}=0\) or \(-6C_{1}=10C_{2}\). Re-arranging, we find the resulting relationship \(3C_{1} = -5C_{2}\).