Energy stored in a capacitor
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How can the energy stored in a capacitor be determined from a potential-charge graph?
The electric potential energy stored in a capacitor is equal to the area under the potential-charge (V-Q) graph. This area represents the work done to charge the capacitor.
State the formula that relates the energy (W) stored in a capacitor to its charge (Q) and potential difference (V).
The energy stored in a capacitor is given by the formula W = (1/2)QV, where W is the energy in Joules, Q is the charge in Coulombs, and V is the potential difference in Volts.
State the formula that relates the energy (W) stored in a capacitor to its capacitance (C) and potential difference (V).
The energy stored in a capacitor can also be calculated using the formula W = (1/2)CV², where W is the energy in Joules, C is the capacitance in Farads, and V is the potential difference in Volts.
A 100μF capacitor is charged to a potential difference of 10V. Calculate the energy stored in the capacitor.
Using the formula W = (1/2)CV², W = (1/2) * (100 × 10⁻⁶ F) * (10 V)² = 0.005 J. The energy stored in the capacitor is 0.005 Joules.
A capacitor stores 0.02 J of energy when charged to 200V. Calculate the charge on the capacitor.
Using the formula W = (1/2)QV, rearrange to find Q = 2W/V. Therefore, Q = (2 * 0.02 J) / 200 V = 2 × 10⁻⁴ C. The charge on the capacitor is 0.2 mC.
How does increasing the potential difference across a capacitor affect the energy stored?
Increasing the potential difference across a capacitor increases the energy stored. Since energy is proportional to V² (W = 1/2 CV²), doubling the voltage quadruples the energy stored.
A parallel-plate capacitor is fully charged and then disconnected from the power supply. If the plate separation is doubled, how does the stored energy change?
Doubling the plate separation halves the capacitance (C=ε₀A/d). Since the charge remains constant (isolated system), energy increases (W=Q²/2C). Halving C, doubles W, resulting in double the energy.
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