Please Help Capacitor Problem electronics?

Question

The question "A capacitor has capacitance C = 6 μF and a charge Q = 2 nC. If the charge is increased to 4 nC what will be the new capacitance?"

The answer in the book is still 6 μF because Capacitance depends on the structure of the capacitor, not on its charge. 
 
Please tell me why, because i know that Q(charge)= Capacitance*Voltage. So why it says it does not depend on charge where it is very clear in the formula? Or the book wrong?

billrussell42 · Accepted Answer

if the capacitor is fixed, it is fixed, and the only way to increase the charge is to increase the voltage.

as it says, capacitor value is dependent only on dimensions and materials.

in this case, Q = CV
V = Q/C = 2nC/6µF = 333 volts
changed to
V = 4nc/6µF = 667 volts

same with resistors, their value is fixed. If you have 100 Ω and 1 amp, you have 100 volts.

if the current increases to 2 amps, you have 200 volts across the (fixed) resistor.

Philomel · Answer

The C is a physical quantity fixed by construction just as a gallon jar is a gallon.
The charge in the capacitor is controlled by the voltage on it. Q=C*V just as the height of the water is controlled by the amount of water in the jar.
Put in more charge and voltage increases.
Pull out charge V goes   v.
There is a variable capacitor but that is another question.
This cap is 6uF !  Given qty in question.

az_lender · Answer

The only way of getting the charge to increase would be by increasing the voltage in the circuit.  The capacitor is a device that will store charge in a ratio Q/V, but that ratio is fixed by the particular capacitor.

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Please Help Capacitor Problem electronics?

3 Answers