Calculating Charge on a Parallel Plate Capacitor with Changing Distance

[jaydnul]

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?

 

[gneill]

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.

 

[jaydnul]

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