As the voltage in the circuit shown below is increased (but not to the breakdown voltage), the capacitance
c. does not change.
d. increases, decreases, or does not change, depending on the charge on the plates of the capacitor.
e. does none of these.
C = Q/V or C = (epsilon_0*A)/(d)
The Attempt at a Solution
If the voltage increases at a specific rate, then the charge would also increase since Q and V are directly proportional and the ratio between the two is a constant. Therefore, the capacitance does not change????