A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 uC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

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To find the potential difference between the plates, use the formula for capacitance: \( C = \frac{Q}{V} \), where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the potential difference. Rearrange the formula to solve for \( V \): \( V = \frac{Q}{C} \). Substitute \( Q = 3.90 \mu C \) and \( C = 920 \text{ pF} \) into the equation.

For part (b), if the plate separation is doubled, the capacitance will change. The capacitance of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. Doubling \( d \) will halve the capacitance, so the new capacitance \( C' = \frac{C}{2} \). Use the formula \( V' = \frac{Q}{C'} \) to find the new potential difference.

For part (c), calculate the work required to double the separation. The work done on a capacitor is given by \( W = \frac{1}{2} Q (V' - V) \), where \( V' \) is the new potential difference and \( V \) is the initial potential difference. Substitute the values of \( Q \), \( V \), and \( V' \) from the previous steps into this formula to find the work done.

Ensure that all units are consistent when performing calculations. Convert microcoulombs to coulombs and picofarads to farads as necessary. Remember that \( 1 \mu C = 1 \times 10^{-6} \text{ C} \) and \( 1 \text{ pF} = 1 \times 10^{-12} \text{ F} \).

Review the concepts of capacitance and potential difference in a parallel-plate capacitor. Understand how changes in physical parameters like plate separation affect the electrical properties of the capacitor, and how energy is involved in these changes.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Capacitance

Capacitance is the ability of a system to store charge per unit voltage, measured in farads (F). For a parallel-plate capacitor, capacitance is determined by the area of the plates, the distance between them, and the permittivity of the material between the plates. It is given by the formula C = Q/V, where C is capacitance, Q is charge, and V is voltage.

Potential Difference

Potential difference, or voltage, is the work done per unit charge to move a charge between two points in an electric field. For capacitors, it is calculated using the formula V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance. It represents the energy required to move charges between the plates.

Work Done in Electric Fields

Work done in electric fields refers to the energy required to move charges within the field. For capacitors, changing the separation between plates affects the electric field and potential energy. The work done to change the separation is calculated using W = 0.5 * Q * ΔV, where W is work, Q is charge, and ΔV is the change in potential difference.