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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 μF, 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 μF 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 connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
An electron is moved from point \(A\), where the electric potential is \(V_{A}=-240 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=-360 \mathrm{V}\). What is the change in the electric potential energy?
The potential difference across a cell membrane from outside to inside is initially at \(-90 \mathrm{mV}\) (when in its resting phase). When a stimulus is applied, Na" ions are allowed to move into the cell such that the potential changes to \(+20 \mathrm{mV}\) for a short amount of time. (a) If the membrane capacitance per unit area is $1 \mu \mathrm{F} / \mathrm{cm}^{2},\( how much charge moves through a membrane of area \)0.05 \mathrm{cm}^{2} ?\( (b) The charge on \)\mathrm{Na}^{+}\( is \)+e$ How many ions move through the membrane?
Two metal spheres have charges of equal magnitude, $3.2 \times 10^{-14} \mathrm{C},$ but opposite sign. If the potential difference between the two spheres is \(4.0 \mathrm{mV},\) what is the capacitance? [Hint: The "plates" are not parallel, but the definition of capacitance holds.]
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
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