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Chapter 17: Problem 79

A parallel plate capacitor is composed of two square plates, $10.0 \mathrm{cm}\( on a side, separated by an air gap of \)0.75 \mathrm{mm} .$ (a) What is the charge on this capacitor when there is a potential difference of \(150 \mathrm{V}\) between the plates? (b) What energy is stored in this capacitor?

Short Answer

Expert verified
Answer: (a) The charge on the capacitor is \(1.771 \times 10^{-9} \ C\). (b) The energy stored in the capacitor is \(1.325 \times 10^{-7} \ J\).

Step by step solution

01

Finding the Capacitance (C)

To find the capacitance, use the formula C = (蔚鈧� * A) / d. The vacuum permittivity (蔚鈧�) is approximately \(8.854 \times 10^{-12} F/m\). The area (A) of a 10.0cm x 10.0cm square plate is \(0.010 \times 0.010 = 0.0001 m^2\). The distance (d) between the plates is given as 0.75mm or \(0.75 \times 10^{-3} m\). Now plug in these values: C = (\((8.854 \times 10^{-12}) \times 0.0001) / (0.75 \times 10^{-3})\) = \(11.805 \times 10^{-12} F\)
02

Finding the Charge (Q)

To find the charge (Q) on the capacitor, use the formula Q = C * V, where C is the capacitance we just determined (11.805 x 10^-12 F) and V is the potential difference given (150 V). Plugging in the values: Q = (\(11.805 \times 10^{-12}) \times 150 = 1.771 \times 10^{-9} C \) (a) The charge on the capacitor is \(1.771 \times 10^{-9} \ C\).
03

Finding the Energy Stored (U)

To find the energy (U) stored in the capacitor, use the formula U = 1/2 * C * V虏, where C is the capacitance (11.805 x 10^-12 F) and V is the potential difference (150 V). Plugging in the values: U = \(0.5 \times (11.805 \times 10^{-12}) \times (150)^2 = 1.325 \times 10^{-7} J\) (b) The energy stored in the capacitor is \(1.325 \times 10^{-7} \ J\).

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Most popular questions from this chapter

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the electric force on the drop is found to be $9.6 \times 10^{-16} \mathrm{N}\( and the potential difference between the plates is \)480 \mathrm{V}$ what is the magnitude of the charge on the drop in terms of the elementary charge \(e\) ? Ignore the small buoyant force on the drop.
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