Three large plates are arranged as shown. How | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Initially, the plates are isolated. The charge on the inner faces of the plates is determined by the distribution of charges $Q$, $2Q$, and $0$ (o

Vedclass · Accepted Answer

none

Vedclass · Answer

(D) Initially, the plates are isolated. The charge on the inner faces of the plates is determined by the distribution of charges $Q$, $2Q$, and $0$ (on the third plate). The charge on the inner face of the first plate is $-Q/2$ and on the left face of the middle plate is $Q/2$. When the switch $K$ is closed, the first and third plates are connected, making them at the same potential. Let the charge on the inner face of the first plate be $-q_1$ and on the left face of the middle plate be $q_1$. Let the charge on the right face of the middle plate be $q_2$ and on the inner face of the third plate be $-q_2$. The total charge on the first and third plates is $Q + 0 = Q$. Thus, $-q_1 - q_2 = Q$, or $q_1 + q_2 = -Q$. The potential difference between the first and middle plate is $V_1 = q_1 d / (\epsilon_0 A)$ and between the middle and third plate is $V_2 = q_2 (2d) / (\epsilon_0 A)$. Since the plates are connected, the potential difference across the capacitors formed must be equal, so $V_1 = V_2$, which implies $q_1 d = q_2 (2d)$, or $q_1 = 2q_2$. Substituting $q_1 = 2q_2$ into $q_1 + q_2 = -Q$, we get $3q_2 = -Q$, so $q_2 = -Q/3$ and $q_1 = -2Q/3$. The charge on the outer face of the first plate is $(Q - (-q_1))/2 = (Q - 2Q/3)/2 = Q/6$. The charge originally on the inner face of the first plate was $-Q/2$. After closing, it is $-q_1 = 2Q/3$. The charge that flows through the key $K$ is the change in charge on the first plate: $\Delta q = q_{final} - q_{initial} = (Q/6 + 2Q/3) - (Q/2) = 5Q/6 - Q/2 = Q/3$. Wait, re-evaluating the flow: The charge on the first plate changes from $Q$ to $Q_{final}$. The charge on the first plate is $Q_{outer} + Q_{inner} = Q/6 + 2Q/3 = 5Q/6$. Change in charge = $5Q/6 - Q = -Q/6$. The magnitude of charge flowing is $Q/6$.

Three large plates are arranged as shown. How much charge will flow through the key $K$ if it is closed?

Similar Questions

Sixty-four conducting drops,each of radius $0.02 \ m$ and each carrying a charge of $5 \ \mu C$,are combined to form a bigger drop. The ratio of the surface charge density of the bigger drop to that of the smaller drop will be ............

In the circuit shown in the figure,$C_1 = C_2 = 2 \mu F$. Find the charge stored in the capacitors.

In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be respectively

$A$ capacitor of capacitance $C$ is connected to a battery of $V$ volts. Now,the distance between the plates of the capacitor is halved while keeping the charge constant,and it is again charged to $V$ volts. What is the energy supplied by the battery?

Vedclass Test Series

Exam Paper Generator

Online Exam Module