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## Homework Statement

A parallel plate capacitor with plates of area [itex]460 cm^2[/itex] is charged to a potential difference [itex]V[/itex] then disconnected from the source of voltage. When the plates move [itex].3 cm[/itex] farther apart, the voltage between them increases by [itex]120 V[/itex]. What is the charge [itex]Q[/itex] on the positive plate of the capacitor? Answer in [itex]nC[/itex]

## Homework Equations

[tex]Q=\frac{ε_0AV}{d}[/tex]

## The Attempt at a Solution

[tex]V=\frac{Qd}{ε_0A}[/tex]

[tex]Q=\frac{ε_0A(V+120)}{(d+.003)}[/tex]

Inserting one into the other...

[tex]Q=\frac{ε_0A}{(d+.003)}(\frac{Qd}{ε_0A}+120)[/tex]

distribute and canceling yields...

[tex]Q=\frac{Qd+120ε_0A}{(d+.003)}[/tex]

Finally

[tex]Q=\frac{120ε_0A}{.003}[/tex]

So using [itex]A=.046 m^2[/itex], I keep getting [itex]16nC[/itex]

Am I doing something wrong assuming [itex]Q[/itex] is constant?