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g100num [7]
3 years ago
12

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i

s produced by charges on the plates. The center plane at x = 0.00 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m. with an initial kinetic energy K = 300 eV, in the positive x-direction.
a. Determine the initial velocity of the electron at x = 0.0 m.
b. What is the potential difference between the plates?
c. At a certain point the electron stops momentarily and then reverses direction. What is the electric potential at this point?
d. What is the kinetic energy of the electron when it reaches plate A?
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

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The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

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Answer:

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3 years ago
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
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The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

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In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

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Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

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