A simple particle accelerator is built using a large parallel plate capacitor (two plates of metal at different voltages separated by some distance) sending a charged particle between them towards a whole in one plate. A diagram of this is below. If the two plates are given a potential V₁ = OV and V = 120V what would the exiting speed of an electron be if it started out stationary at the Vį plate? Vi Vf
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A: The potential difference V=V1-V2=825-125=700 Volts
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A: C1 = 3.10 μF,C2 = 4.50 μF, andVab = +63.0 V.
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- You then fill the capacitor with a dielectric material of dielectric constant κκ while the capacitor remains connected to the same potential difference. What is the new charge on the plates? What is the new energy stored by the capacitor?#29: In Figure P21.29, the electric potential at point A is -300 V. What is the potential at point B, which is 5.0 cm to the right of A?. 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
- the selected answer is correct, I did that randomly. what's the process of getting here?A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 400 V. 1) If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate? (Express your answer to two significant figures.)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…
- In the figure (Figure 1), C1C1 = C5C5 = 8.1 μFμF and C2C2= C3C3 = C4C4 = 5.0 μFμF . The applied potential is VabVab = 200 VV .Calculate the charge on capacitor C2C2.Calculate the potential difference across capacitor C2C2.Calculate the charge on capacitor C3C3.Calculate the potential difference across capacitor C3C3.Calculate the charge on capacitor C4C4.Calculate the potential difference across capacitor C4C4.Calculate the charge on capacitor C5C5.Calculate the potential difference across capacitor C5C5.A 10.0-µF capacitor is charged so that the potential difference between its plates is 10. V. A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V. The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate. What is the final potential difference across the plates of the capacitors when they are connected in parallel? O 8.3 V O 6.7 V O 10 V O 7.5 V O 5.0VIn the figure, a proton is fired with a speed of 200,000 m/s from the midpoint of the capacitor toward the positive plate. a. Show that this is insufficient speed to reach the positive plate. b. What is the proton’s speed as it collides with the negative plate?
- An electron passes through an area of changing potential as shown below. At point A, the electron has a speed of 7.2 × 106 m/s. What is the approximate speed of the electron at point B? A +10 V 7 +20 V B +30 VProblem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is d₂ = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of V₂ = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vf between the plates after the plates have been pulled to their new, larger separation df. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will…This is a part of a review I'm studying, NOT a graded assignment, please do not reject.