Answered: An electronic flash for a camera uses a… | bartleby

Related questions

Question

2. An electronic flash for a camera uses a capacitor to store energy. With a potential difference of
300V, the charge on each plate has a magnitude of 0.0225C
What is the capacitance of the flash?
If this is a parallel plate capacitor of area 10m?, what is the plate separation?
How much energy is stored by the capacitor?
а.
b.
С.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A parallel-plate capacitor of capacitance C0 = ε0A0/d0 , is connected to a battery. When the capacitor is fully charged, it is disconnected from the battery (initial state). We push the plates closer together, until d = d0/2 (final state). a) How does the energy stored in the capacitor change (increase or decrease)? Justify your answer. b) Calculate Ufinal / Uinitial , in terms of the parameters defined in the problem.
. A parallel-plate capacitor is made of two circular plates, each with a diameter of 2.50x10^-3 m. The plates of this capacitor are separated by a space of 1.40x10^-4 m. How much charge will be stored on each plate of this capacitor when it is connected across a potential difference of 0.12V? A. 3.7x10^-8 µC B. 4.2x10^-8 µC C. 5.6x10^-8 µC D. 6.1x10^-8 µC
A parallel-plate capacitor is made of two square plates 25 cm on a side and 1.5 mm apart. The capacitor is connected to a 70-V battery. Hint a. What is the energy stored in the capacitor? Energy stored in the capacitor is b. With the battery still connected, the plates are pulled apart to a separation of 3 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is HJ. µJ. c. This time, starting from situation in (a), with the batteries disconnected (but capacitors still charged), the plates are pulled apart to a separation of 3 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is d. Comparing your results in (b) and (c) above, it makes sense that the energy stored in the capacitor increases in (c), because the work done in separating the plates is stored as the electrostatic potential energy. In (b), why does the energy decrease even though work is done in separating the plates? HJ. O A negative work is done in…
Consider a parallel-plate capacitor made up of two conducting plates with dimensions 34 mm × 18 mm. a) If the separation between the plates is 1.1 mm, what is the capacitance, in pF, between them? b) If there is 0.32 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant?
A parallel plate capacitor is made from two aluminum foil sheets, each 3.8cm wide and 6.1 m long. Between the sheets is a Teflon strip of the same width and length that is 0.025mm thick. What is the energy stored in the capacitor when the charge on the capacitor is 30 uC?
Charge q is fired through a small hole in the positive plate of a capacitor, as shown.a. If q is a positive charge, does it speed up or slow down inside the capacitor? Answer this question twice: (i) Using the concept of force. (ii) Using the concept of energy.b. Repeat part a if q is a negative charge.
A parallel-plate capacitor is made of two square plates 20 cm on a side and 1 mm apart. The capacitor is connected to a 50-V battery. Hint # a. What is the energy stored in the capacitor? Energy stored in the capacitor is b. With the battery still connected, the plates are pulled apart to a separation of 2 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is c. This time, starting from situation in (a), with the batteries disconnected (but capacitors still charged), the plates are pulled apart to a separation of 2 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is d. Comparing your results in (b) and (c) above, it makes sense that the energy stored in the capacitor- increases in (c), because the work done in separating the plates is stored as the electrostatic potential energy. In (b), why does the energy decrease even though work is done in separating the plates? E D Submit Question с 4 O The energy is not…
An air-filled parallel-plate capacitor has plates of area 2.50 cm^2 separated by 1.80 mm. The capacitor is connected to a(n) 13.0 V battery. a) Find the value of its capacitance. pF b) What is the charge on the capacitor? pC c) What is the magnitude of the uniform electric field between the plates? N/C
A capacitor of 3.23µF has an area of 6.35mm^2. Determine the separation distance between the two plates. If the new capacitance is now 5.63mF, what is the value of the dielectric inserted to the capacitor? What is the new voltage if the capacitor is connected to a 110 volts source? If the electric field created by two plates 8.99x10^4 N/C and a working voltage of 63 volts. Will the capacitor experience a dielectric breakdown? Please prove your answer. Note: This question is under the topic "Capacitance". Thank you!