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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 = (ε₀ * A) / d. The vacuum permittivity (ε₀) 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 * V², 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

Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of \(0.80 \mu \mathrm{C}\) on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?
The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?
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