# Capacitor change in distance

## Homework Statement

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

## Homework Equations

$$Q=\frac{ε_0AV}{d}$$

## The Attempt at a Solution

$$V=\frac{Qd}{ε_0A}$$
$$Q=\frac{ε_0A(V+120)}{(d+.003)}$$
Inserting one into the other...
$$Q=\frac{ε_0A}{(d+.003)}(\frac{Qd}{ε_0A}+120)$$
distribute and canceling yields...
$$Q=\frac{Qd+120ε_0A}{(d+.003)}$$
Finally
$$Q=\frac{120ε_0A}{.003}$$

So using $A=.046 m^2$, I keep getting $16nC$

Am I doing something wrong assuming $Q$ is constant?

## Answers and Replies

gneill
Mentor
Your assumption about Q being constant is correct, and I don't see any errors in your method. That leaves significant figures or application errors (the marking program has been given a wrong answer to look for) as suspects.

Ok thanks for the feedback. This homework system has a reputation for making errors.