Question

A particle has a charge of +1.0 μC and moves from point A to point B, a distance of 0.14 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.3 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

Answer 1

(a) F =
`5.92* 10^(- 3)\ N`

(b) E = 5920 N/C

Solution:

As per the question:

Charge, q =
`+ 1.0\ mu C = + 1.0* 10^(- 6)\ C`

`\Delta U = EPE_(A) - EPE_(B) = + 8.3* 10^(- 4)\ J`

Distance, d = 0.14 m

Now,

(a) The magnitude of the electric force on the particle:

Work done, W =
`\Delta U`

W = Fd

where

F = Force

F =
`(\Delta U)/(d)`

F =
`(8.3* 10^(- 4))/(0.14) = 5.92* 10^(- 3)\ N`

(b) The magnitude of the electric field:

`F = qE`

`E = (F)/(q) = (5.92* 10^(- 3))/(1.0* 10^(- 6)) = 5920\ N/C`