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The elctric field between two metal plates is assumed to be uinform at 2*10^4V/m.A positively charged particle has displacement of 0.2m in the direction of the field .What is the change in PE if it carries a charge of +3e?e=1.6*10^-19C.

The  elctric field between two metal plates is assumed to be uinform at 2*10^4V/m.A positively charged particle has displacement of 0.2m in the direction of the field .What is the change in PE if it carries a charge of +3e?e=1.6*10^-19C.

Grade:Select Grade

1 Answers

Shivam Chopra
45 Points
6 years ago
Electric field b/w the plates = σ/ϵ = 2 * 104 Vm-1
Charge on particle = +3e = 4.8 * 10-19 C
So, force on particle = 4.8 * 2 * 10-15 = 9.6 * 10-15N
As, the particle moves in the direction of field, so, work is done by the system. So, potential energy will be –ve.
So, PE = - F * d = - 9.6*10-15 * 0.2 =  - 1.92 * 10-15 J.
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