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, we need to consider the formula for the energy stored in a capacitor:
E = (1/2) * C * V^2,
where E represents the energy stored, C is the capacitance of the capacitor, and V is the potential difference across the capacitor.
(a) If the potential difference (V) across the capacitor is doubled, we can find the new energy stored by plugging the new value of V into the energy formula:
E' = (1/2) * C * (2V)^2,
Simplifying this equation gives us:
E' = (1/2) * C * 4V^2,
E' = (1/2) * 4C * V^2,
E' = 2 * (1/2) * C * V^2,
E' = 2E.
From this calculation, we can see that when the potential difference across the capacitor is doubled, the energy stored in the capacitor also doubles.
(b) If the charge on each plate of the capacitor is doubled and the capacitor remains connected to a battery, we need to relate the charge, potential difference, and capacitance. The formula to connect these variables is:
Q = C * V,
where Q is the charge on each plate, C is the capacitance, and V is the potential difference.
If the charge doubles, we can write the new charge as 2Q. Substituting this into the equation above, we get:
2Q = C * V',
Here, V' represents the new potential difference.
Rearranging the equation, we can solve for the new potential difference:
V' = (2Q) / C.
We can now substitute this value into the energy formula:
E' = (1/2) * C * (V')^2,
E' = (1/2) * C * [(2Q) / C]^2,
E' = (1/2) * (4Q^2 / C),
Simplifying this expression gives us:
E' = 2 * (1/2) * (Q^2 / C),
E' = Q^2 / C.
From this calculation, we can see that when the charge on each plate of the capacitor is doubled, the energy stored in the capacitor increases by a factor of 4 (Q^2 / C compared to E).