Two oppositely charged, parallel plates are separated by 5.26 mm. A potential di

Question

Two oppositely charged, parallel plates are separated by 5.26 mm. A potential difference of 600. V exists between the plates.
NOTE: You must enter units for each of the answers below.

(a) Calculate the magnitude of the electric field strength between the plates.

(b) Calculate the magnitude of the force on an electron between the plates.

(c) An electron is moved from the negative plate to the positive plate. What is the change in electric potential energy of the electron? (Think about whether the electron's potential energy will increase or decrease in this process.)

(d) Calculate change in electric potential energy of the electron if it is initially positioned 2.95 mm from the positive plate and moved to the negative plate. (Again, think about whether the electron's potential energy will increase or decrease in this process.)

Explanation / Answer

d = 5.25 mm
V = 600 V
(a)
We know, E = V/d
E = 600/(5.25 * 10^-3)
E = 1.14 * 10^5 V/m

(b)
F = q*E
F = 1.6 * 10^-19 * 1.14 * 10^5 N
F = 1.824 * 10^-14 N

(c)
Change in eletric potential energy = q*V
Change in eletric potential energy = 1.6 * 10^-19 * 600 J
Change in eletric potential energy = 9.6 * 10^-17 J

Potential energy will decrease

(d)
Potential at that point, V = E.d = 1.14 * 10^5 * 2.95 * 10^-3 = 336.3 V
Change in eletric potential energy = q*V
Change in eletric potential energy = 1.6 * 10^-19 * 336.3 V
Change in eletric potential energy = 5.38 * 10^-17 J

Potential energy will increase

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