Question

Four very large metal plates are given charges as shown. The middle two are then connected through a conducting wire. The charge that will flow through the wire is

• A. Q
• B. 2Q
• C. 3Q
• D. 0

Answer 1

5Q

Answer 2

To solve this problem, we need to understand how charges distribute on large conducting plates and how they redistribute when conductors are connected.

1. Initial Charge Distribution

Let the four plates be P1, P2, P3, P4 from left to right, with initial charges $Q_1 = Q$, $Q_2 = 2Q$, $Q_3 = -3Q$, $Q_4 = -4Q$.
For very large parallel plates, the charges on the outermost surfaces (left surface of P1 and right surface of P4) are equal and sum up to the total charge of the system.
Total charge $Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 = Q + 2Q - 3Q - 4Q = -4Q$.
So, the charge on the left surface of P1 ($Q_{1L}$) and the right surface of P4 ($Q_{4R}$) is:
$Q_{1L} = Q_{4R} = \frac{Q_{total}}{2} = \frac{-4Q}{2} = -2Q$.

Now, let's find the charges on all surfaces in the initial state. For any adjacent plates, the charges on their facing surfaces are equal in magnitude and opposite in sign. Also, the sum of charges on the two surfaces of a single plate equals its total charge.

• Plate P1:
  $Q_1 = Q_{1L} + Q_{1R} \Rightarrow Q = -2Q + Q_{1R} \Rightarrow Q_{1R} = 3Q$.
  Since P1R faces P2L, $Q_{2L} = -Q_{1R} = -3Q$.

• Plate P4:
  $Q_4 = Q_{4L} + Q_{4R} \Rightarrow -4Q = Q_{4L} + (-2Q) \Rightarrow Q_{4L} = -2Q$.
  Since P4L faces P3R, $Q_{3R} = -Q_{4L} = 2Q$.

• Plate P2:
  $Q_2 = Q_{2L} + Q_{2R} \Rightarrow 2Q = -3Q + Q_{2R} \Rightarrow Q_{2R} = 5Q$.
  Since P2R faces P3L, $Q_{3L} = -Q_{2R} = -5Q$.

• Plate P3 (check):
  $Q_3 = Q_{3L} + Q_{3R} = -5Q + 2Q = -3Q$. This matches the initial charge on P3.

Summary of Initial Surface Charges:

• P1: Left surface = -2Q, Right surface = 3Q
• P2: Left surface = -3Q, Right surface = 5Q
• P3: Left surface = -5Q, Right surface = 2Q
• P4: Left surface = -2Q, Right surface = -2Q

2. Final Charge Distribution after Connecting P2 and P3

When P2 and P3 are connected by a conducting wire, they form a single conductor (let's call it P23). The total charge on this combined conductor is conserved:
$Q_{23, total} = Q_2 + Q_3 = 2Q + (-3Q) = -Q$.

The charges on the outermost surfaces of the entire system (P1L and P4R) remain unchanged.
$Q'_{1L} = -2Q$ and $Q'_{4R} = -2Q$.

Now, let's find the charges on the surfaces of the combined P23 conductor.

• Plate P1:
  $Q_1 = Q'_{1L} + Q'_{1R} \Rightarrow Q = -2Q + Q'_{1R} \Rightarrow Q'_{1R} = 3Q$.
  Since P1R faces P23L (the left surface of the combined conductor), $Q'_{23L} = -Q'_{1R} = -3Q$.

• Plate P4:
  $Q_4 = Q'_{4L} + Q'_{4R} \Rightarrow -4Q = Q'_{4L} + (-2Q) \Rightarrow Q'_{4L} = -2Q$.
  Since P4L faces P23R (the right surface of the combined conductor), $Q'_{23R} = -Q'_{4L} = 2Q$.

• Combined Plate P23 (check):
  The total charge on P23 should be $Q'_{23L} + Q'_{23R} = -3Q + 2Q = -Q$. This matches $Q_{23, total}$.

Summary of Final Surface Charges:

• P1: Left surface = -2Q, Right surface = 3Q
• P23 (combined): Left surface = -3Q, Right surface = 2Q
• P4: Left surface = -2Q, Right surface = -2Q

3. Calculate the Charge Flow

The charge that flows through the wire is the amount of charge that moves from one plate to the other to establish the new equilibrium distribution.
Consider the original Plate 2. Initially, its total charge was $2Q$.
After connection, Plate 2 and Plate 3 become a single conductor. The charge on the left surface of this combined conductor is $Q'_{23L} = -3Q$, which is on the part that was originally Plate 2's left surface. The charge on the right surface of the combined conductor is $Q'_{23R} = 2Q$, which is on the part that was originally Plate 3's right surface.
The charges that were initially on the facing surfaces of P2 and P3 ($Q_{2R}=5Q$ and $Q_{3L}=-5Q$) redistribute internally within the new combined conductor.

Let's consider the charge on the original Plate 2.
Initial charge on Plate 2 = $2Q$.
In the final state, the charge on the left surface of the original Plate 2 is $Q'_{23L} = -3Q$.
The remaining charge of the combined conductor, $Q_{23, total} - Q'_{23L} = -Q - (-3Q) = 2Q$, is distributed on the right surface of the combined conductor. This charge $2Q$ is on the part that was originally Plate 3's right surface.

To find the charge flow, we can consider the change in charge on one of the plates that was connected. Let's consider Plate 2.
Initially, Plate 2 had a total charge of $2Q$.
In the final state, the charge on the portion of the combined conductor that was originally Plate 2 is not simply its final surface charges. Instead, we consider the charge on its "outer" surface (the one facing P1) and the charge that moved to/from it.

A more direct way:
Consider the total charge on the system (P2+P3) initially: $Q_2+Q_3 = 2Q - 3Q = -Q$.
This total charge is conserved.
The charge that flows is the internal redistribution.
Let's look at the charge on the right surface of P2 initially, $Q_{2R} = 5Q$.
Let's look at the charge on the left surface of P3 initially, $Q_{3L} = -5Q$.
When they are connected, these two surfaces essentially merge into the bulk of the new conductor. The charges on these surfaces, $5Q$ and $-5Q$, will neutralize each other and distribute to the new external surfaces of the combined conductor.
The "internal" charge that was on $P_2$'s right surface ($5Q$) effectively flows to $P_3$'s left surface (which had $-5Q$) and then some charge flows to redistribute itself.

Let's re-evaluate the flow from the perspective of one plate.
Initial charge on Plate 2 = $2Q$.
Initial charge on Plate 3 = $-3Q$.

When connected, they form a single body.
The left surface of this body (originally P2's left surface) has charge $-3Q$.
The right surface of this body (originally P3's right surface) has charge $2Q$.

Consider the charge that was originally on Plate 2. It was $2Q$.
After connection, the charge on the left surface of Plate 2 is still $-3Q$.
The "remaining" charge of Plate 2, which was $2Q - (-3Q) = 5Q$, was on its right surface. This $5Q$ charge (or part of it) must flow into the connection to combine with the charge from Plate 3.

Let $q$ be the charge that flows from Plate 2 to Plate 3.
Final charge on Plate 2 = $2Q - q$.
Final charge on Plate 3 = $-3Q + q$.

However, this doesn't directly yield the surface charges.
A better approach is to consider the charge on the "inner" surfaces of the two plates that get connected.
Initially, the right surface of P2 had $Q_{2R} = 5Q$.
Initially, the left surface of P3 had $Q_{3L} = -5Q$.
When connected, these two surfaces are no longer distinct interfaces; they become part of the interior of the single conductor. The charge that was on these surfaces will redistribute.

The charge on the right surface of P2 was $5Q$. This charge effectively flows towards P3 (or neutralizes with charge from P3).
The charge on the left surface of P3 was $-5Q$. This charge effectively flows towards P2 (or neutralizes with charge from P2).
The net flow is the difference between the initial charge on one of the plates and its effective charge after connection, considering the new surface distributions.

Let's assume the charge on P2 changes from $2Q$ to $Q'_2$ and on P3 from $-3Q$ to $Q'_3$.
The new configuration means P2 and P3 are effectively one plate.
The charge on the left side of the combined P23 is $-3Q$. This charge effectively resides on the part that was originally P2.
The charge on the right side of the combined P23 is $2Q$. This charge effectively resides on the part that was originally P3.
The total charge on the combined body is $-Q$.

Consider the charge of P2. Initially, it was $2Q$.
After connection, the charge on its left face ($Q_{2L}$) remains $-3Q$.
The charge on its right face ($Q_{2R}$) was $5Q$. This charge has now redistributed.
The total charge of the combined P23 system is $-Q$.
The charge that flowed is the amount needed to equalize the potential and redistribute the charges.

Let's consider the charge on the left surface of P2 (facing P1) and the right surface of P3 (facing P4).
Initial: $Q_{2L} = -3Q$, $Q_{3R} = 2Q$.
Final: $Q'_{23L} = -3Q$, $Q'_{23R} = 2Q$.
These charges do not change.

The charges that were initially on the inner surfaces are $Q_{2R} = 5Q$ and $Q_{3L} = -5Q$.
These are the charges that redistribute through the wire to form the new combined body.
When connected, the region between P2 and P3 is effectively shorted.
The charge that flows is the amount that moves to neutralize the internal field and redistribute the charge.

Consider the charge that was initially on the right surface of Plate 2, which is $5Q$. This charge effectively moves through the wire to the left surface of Plate 3, which had $-5Q$.
This is not exactly the flow. The flow is the net charge that passes from one side to the other.

Let $q_{flow}$ be the charge flowing from P2 to P3.
Initial charge on P2 = $2Q$. Initial charge on P3 = $-3Q$.
Final charge on P2 + Final charge on P3 = $-Q$.
The final charge distribution on the combined P23 system is $-3Q$ on the left side (originally P2's left) and $2Q$ on the right side (originally P3's right).
The charge of the original P2 is now effectively $-3Q$ (its left surface) plus some part of the "bulk" charge.
The charge of the original P3 is now effectively $2Q$ (its right surface) plus some part of the "bulk" charge.

The easiest way to calculate the charge flow is to consider the charge that moves from one plate to the other.
Let's assume that after connection, the charge on the left part of the combined conductor (which was Plate 2) is $Q'_2$ and on the right part (which was Plate 3) is $Q'_3$.
This is not how it works. When connected, they become a single conductor.

The charge that flows through the wire is the amount of charge that moves from one plate to the other to establish the new equilibrium.
Consider the initial distribution $Q_{2R} = 5Q$ and $Q_{3L} = -5Q$.
When connected, these charges effectively cancel out through the wire. This implies $5Q$ flowed from P2 to P3.
Let's verify this by checking the total charge on P2 if $5Q$ flowed out of it.
Initial charge on P2 = $2Q$.
If $5Q$ flows out, final charge on P2 would be $2Q - 5Q = -3Q$.
This $-3Q$ is exactly the charge found on the left surface of the combined P23 conductor ($Q'_{23L} = -3Q$).
This means that all the charge that was initially on the right surface of P2 ($5Q$) flowed through the wire to P3, leaving P2 with only its left surface charge.
Let's check P3.
Initial charge on P3 = $-3Q$.
If $5Q$ flows into P3, final charge on P3 would be $-3Q + 5Q = 2Q$.
This $2Q$ is exactly the charge found on the right surface of the combined P23 conductor ($Q'_{23R} = 2Q$).

So, the charge that flows through the wire is $5Q$. This corresponds to the charge that was on the inner surface of Plate 2, which flowed to neutralize the charge on the inner surface of Plate 3, and establish the new equilibrium.

Final check:
Initial state:
P1: (-2Q, 3Q)
P2: (-3Q, 5Q) -> Total 2Q
P3: (-5Q, 2Q) -> Total -3Q
P4: (-2Q, -2Q)

When P2 and P3 are connected, $5Q$ from P2's right surface flows to P3's left surface.
P2 now effectively has only its left surface charge: $-3Q$.
P3 now effectively has its right surface charge $2Q$ plus the $5Q$ it received, making it $7Q$. But this is not how it works.

The combined charge of P2 and P3 is $-Q$.
The final surface charges are $Q'_{23L} = -3Q$ and $Q'_{23R} = 2Q$.
The change in charge on Plate 2 is $Q'_{23L} - Q_2 = -3Q - 2Q = -5Q$. This means $5Q$ flowed out of Plate 2.
The change in charge on Plate 3 is $Q'_{23R} - Q_3 = 2Q - (-3Q) = 5Q$. This means $5Q$ flowed into Plate 3.
Therefore, the charge that flowed through the wire is $5Q$.