0Two large, parallel, metal plates carry opposite charges of equal magnitude. Th

Question

0Two 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 375 V .

a)What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?

b)What is the magnitude of the force this field exerts on a particle with a charge of 2.60 nC ?

c)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.

d)Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Explanation / Answer

Here ,

distance , d= 50 mm

potential difference , V = 375 V

a)

electric field in the region = V/d

electric field in the region = 375/(50 *10^-3)

electric field in the region = 7500 V/m

the electric field in the region is 7500 V/m

b)

magnitude of electric force = charge * electric field

magnitude of electric force = 7500 * 2.6 *10^-9 N

magnitude of electric force = 1.95 *10^-5 N

the magnitude of electric force is 1.95 *10^-5 N

c)

work done by the field = - electric force * distance

work done by the field = -1.95 *10^-5 * 0.050

work done by the field = -9.75 *10^-7 J

d)

change in potential energy of charge = - work done by electric field

change in potential energy of charge = 9.75 *10^-7 J

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.