How does the energy stored in a capacitor change if (a) the potential difference is
doubled, and (b) the charge on each plate is doubled, as the capacitor remains connected to a battery?
To understand how the energy stored in a capacitor changes in different scenarios, let's consider each case separately:
(a) If the potential difference (voltage) across a capacitor is doubled, the energy stored in the capacitor will increase. The relationship between the energy stored (E), capacitance (C), and the potential difference (V) is given by the formula: E = 1/2 * C * V².
When the potential difference is doubled, the energy stored in the capacitor will be four times its previous value. This can be derived by substituting the new value of V in the formula. For example, if the potential difference was initially V1 and is now doubled to V2 (where V2 = 2V1), the new energy stored (E2) would be four times the original energy stored (E1).
(b) If the charge on each plate of the capacitor is doubled while the capacitor remains connected to a battery, the energy stored in the capacitor will increase by a factor of four. The relationship between the energy stored (E) in a capacitor, capacitance (C), and charge (Q) is given by the formula: E = 1/2 * C * Q².
When the charge on each plate is doubled, the new energy stored (E2) will be four times the original energy stored (E1). This can be derived by substituting the new value of Q in the formula. For example, if the charge on each plate was initially Q1 and is now doubled to Q2 (where Q2 = 2Q1), the new energy stored (E2) would be four times the original energy stored (E1).
In summary, doubling the potential difference across a capacitor will result in a four-fold increase in energy, while doubling the charge on each plate will also result in a four-fold increase in energy.