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Chapter 17: Problem 81

Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?

Short Answer

Expert verified
Answer: The initial potential difference is 20 V, and the initial charge is 0.002 C or 2.0 mC.

Step by step solution

01

Determine the Initial Potential Difference

First, let's find the initial potential difference (V) using the average power equation. Average power (P_avg) = 0.5 * C * V^2 / time We are given: C = 100.0 渭F, P_avg = 10.0 kW, and time = 2.0 ms. To solve for V, we rearrange the formula: V^2 = (2 * P_avg * time) / C Now, plug in the given values and convert units: V^2 = (2 * (10.0 * 10^3 W) * (2.0 * 10^-3 s)) / (100.0 * 10^-6 F) V^2 = (20,000 * 0.002) / 0.0001 V^2 = 400 Take the square root of both sides: V = 20 V So, the initial potential difference is 20 V.
02

Determine the Initial Charge

Next, let's find the initial charge (Q) using the formula: Charge (Q) = Capacitance (C) * Potential difference (V) We are given: C = 100.0 渭F and V = 20 V. Plug in the given values and convert units: Q = (100.0 * 10^-6 F) * (20 V) Q = 0.002 C So, the initial charge is 0.002 C or 2.0 mC.

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Most popular questions from this chapter

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?
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 tiny hole is made in the center of the negatively and positively charged plates of a capacitor, allowing a beam of electrons to pass through and emerge from the far side. If \(40.0 \mathrm{V}\) are applied across the capacitor plates and the electrons enter through the hole in the negatively charged plate with a speed of \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) what is the speed of the electrons as they emerge from the hole in the positive plate?
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
By rewriting each unit in terms of kilograms, meters, seconds, and coulombs, show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\)
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