Question

Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?

Double the charge and double the plate area.
Double the charge and double the plate separation.
Halve the charge and double the plate separation.
Halve the charge and double the plate area.
Halve the plate separation and double the plate area.
Double the plate separation and halve the plate area.

Answer 1

Halve the plate separation and double the plate area

Step-by-step explanation:

Since capacitance is given by
`C=(\epsilon_(o) A)/(d)`

Where A is the plate area and d is the plate separation, decreasing the separation by 0.5 and doubling the area will make the capacitance increase 4 times if
`\epsilon_(o)` is constant. Therefore, Halve the plate separation and double the plate area