In the middle of two plates making a parallel plate capacitor a plate was placed, having charge q. The plate is moved parallel to itself by a distance x. What is the charge (in μ C ) flowing through t

Question

In the middle of two plates making a parallel plate capacitor a plate was placed, having charge q. The plate is moved parallel to itself by a distance x. What is the charge (in $μ C$ ) flowing through the external circuit of the capacitor if the distance between its plates equal to d ? Take $q = 8 μ C, x = 1 mm, d = 4 mm$ .

TestHub · Accepted Answer

Let the middle plate carry charge q; it induces -q1 and -(q-q1) on the inner faces of the outer plates.

The two outer plates are connected, so they are at the same potential: V1 = V2.

$\frac{q _{1}}{A ε _{0}} \times 2 = \frac{q - q _{1}}{A ε _{0}} \times 2 \Rightarrow q_{1} = \frac{q}{2}$

When the middle plate shifts by x = 1 mm, the new charge distribution gives $q_{1}^{'} = \frac{3 q}{4}$ (with d = 4 mm).

$Δ q = q_{1}^{'} - q_{1} = \frac{3 q}{4} - \frac{q}{2} = \frac{q}{4} = \frac{8}{4} = 2 μ C$

Answer: 2.00

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