Answer to Question #302089 in Electric Circuits for Natz

Question

Infer what will happen to the capacitance, charge, and potential difference in the respective capacitors when their size, potential difference, charge will change?

Answer 1

The capacitance is related to the charge and potential difference as $C = \dfrac{q}{U}$ . Also we may write a formula to express the capacitance using the geometric parameters S (area of plates) and d (distance between the plates) and the permittivity $\varepsilon$ : $C = \dfrac{\varepsilon\varepsilon_0 S}{d}.$

1. Let us determine the change in capacitance. If the area increases, the capacitance also increases proportionally. If the distance d increases, the capacitance decreases. If we want to increase potential difference U and save the charge q constant, we should decrease the capacitance. If we want to increase the charge and retain the potential difference constant, we should increase the capacitance.
2. Let us determine the change in charge. If the area of plates increases and other parameters remain constant, we may increase the stored charge. If the distance d increases, the capacitance decreases and the charge will also decrease. If the potential difference increases and other parameters remain the same, the charge will increase.
3. Let us determine the change in potential difference. If the area increases and other parameters remain the same, the capacitance increases and for the constant charge the potential difference should decrease. If the distance d increases, the capacitance decreases and for the constant charge the potential difference should increase. If the charge increases and the capacitance remains the same, the potential difference should increase.