A proton enters a pair of charged parallel plates (which we call a capacitor) traveling to the right at a speed of 1.276 10-5 m/s, as shown below. The separation of the plates is 1.61 cm. The proton enters the capacitor midway between the plates; that is, a distance r = 0.805 cm from each plate, as shown. The capacitor generates a 2.90 10-4 N/C uniform electric field between the plates, pointing downward. What horizontal distance does the proton traverse before the proton hits the bottom plate? [Assume gravity does NOT play a significant role.] m

Q: A parallel-plate vacuum capacitor has 6.88 J of energy stored in it. The separation between the…

A: Capacitance of a parallel plate capacitor When the separation is decreased, the capacitance…

Q: A parallel-plate capacitor has a capacitance of 500 pF, a plate area of 550 cm2, and a mica…

A: Given  C, capacitance = 500 pF  Plate area, A = 550 cm2  Dielectric constant, k = 5.40  Potential, V…

Q: Part A If the separation is decreased to 1.85 mm, what is the energy now stored if the capacitor was…

A: Given a parallel plate capacitor which has U0 =7.42J of energy stored in it and the separation of…

Q: A parallel plate capacitor, in which the space between the plates is empty, has a capacitance of Co=…

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Q: A 68.3-V potential difference is applied across the parallel plates of a 0.15-μF capacitor. (a) What…

A:

Q: A capacitor is constructed of two parallel plates of side l separated by a distance d. A material of…

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Q: A parallel-plate capacitor has a capacitance of 680 pF, a plate area of 580 cm², and a mica…

A: Given data, Capacitance C=680 pF Dielectric K=5.4 Voltage V=26 V Area=580 cm2

Q: A parallel-plate capacitor has a capacitance of 260 pF, a plate area of 170 cm², and a mica…

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Q: A parallel plate capacitor with capacitance C=0.6 mF is connected and charged to a battery with…

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Q: Consider a round parallel plate capacitor consisting of two circular metal plates of radius, R = 1.0…

A: To Calculate the capacitance of this capacitor we will use the formula of capacitance as given by ,…

Q: A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the…

A: We know that ,Energy stored in the capacitor is, U=Q2d2ε0Awhere charge on the plates remains…

Q: With vacuum between its plates, a parallel-plate capacitor has capacitance 5.60 mF. You attach a…

A: given,C=5.6mFa)U=12CV2=125.6×10-6×1.5×1032=6.3J

Q: The two identical square plates of a parallel plate capacitor in vacuum have a separation distance…

A: Given, separation between the plates of parallel plate capacitor, d = 0.050 mm Capacitance of…

Q: A proton enters the uniform electric field produced by the two charged plates shown below. The…

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Q: Suppose the particle is placed at the 40V equipotential surface along the line connecting the two…

A: Given information: The charge of the particle, q=80 μC=80×10-6 C The potential at different points.…

Q: A 5.30 cm by 2.40 cm parallel plate capacitor has the plates separated by a distance of 2.00 mm.…

A: The objective of the question is to find the electric field in the dielectric of a parallel plate…

Q: A parallel-plate capacitor has a capacitance of 680 pF, a plate area of 580 cm², and a mică…

A: Given that: The capacitance is C=680 pF= 680×10-12 F The area of the plate is A=580 cm2=580×10-4 m2…

Q: ASK YOUR TEACHER PRACTICE ANOTHER A parallel plate capacitor has plates of area A = 5.00 ✕ 10−2 m2…

A: Given:  Parallel plate capacitor's area, A=5×10-2 m2 Separation between plates, d=1.47×10-4 m…

Q: A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the…

A: Since the work done for both the charges to move from one plate to another is same.

Q: A parallel-plate vacuum capacitor has 8.08 J of energy stored in it. The separation between the…

A:

Q: f the separation is decreased to 1.15 mm, what is the energy now stored if the capacitor was…

A: Given that Initial energy stored is U0 = 8 J separation between the plates is d = 3.20 mm

Q: A parallel-plate capacitor has a capacitance of 400 pF, a plate area of 810 cm², and a mica…

A: Given data: Capacitance of the capacitor (C) = 400 pF Plate area (A) = 810 cm2  Dielectric constant…

Q: alculate the capacitance (in F) if the space between the plates is filled with air. the capacitance…

A: a) We know that the formula of the capacitance if the space between the plate is filled with air is…

Q: A parallel-plate capacitor has a capacitance of 630 pF, a plate area of 420 cm?, and a mica…

A: Given                Capacitance,  C= 630 pF                Area of plates A= 420 cm2…

Q: A capacitor is constructed out of two circular plates, each of radius 6 cm. The plates are…

A: Radius of each capacitor, r=6 cm=6×10-2 m Thus area of each plate, A=πr2=π×(6×10-2)2 m2 Separation…

Q: Problem 12: A capacitor consists of two parallal planar plates, each plate having an area of 16.0m?.…

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Q: A parallel-plate capacitor initially has Quartz between its plates and carries charge ±Q. A…

A: The dielectric constant (k) of the new material should be approximately 6.25. Explanation:The energy…

Q: A parallel-plate capacitor with vacuum between the two plates has the plate area (of one plate)…

A: The capacitance (C) of a parallel-plate capacitor with vacuum between the plates can be calculated…

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.