Two large, parallel, metal plates carry opposite charges of equal magnitude. The

Question

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 50.0 mm, and the potential difference between them is 350 V.

What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
I have 7000 V/M but I need it in N/C..

What is the magnitude of the force this field exerts on a particle with a charge of 2.30 nC?
(in Newtons)
Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
(in Joules)
Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
(in Joules)

Explanation / Answer

E =V/d = 350/50*10-3 = 7000 V/m

F= Eq = 7000*2.3*10-9 =1.61*10-5 N

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