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

Answer:

(a) F = $5.92\times 10^{- 3}\ N$

(b) E = 5920 N/C

Solution:

As per the question:

Charge, q = $+ 1.0\ mu C = + 1.0\times 10^{- 6}\ C$

$\Delta U = EPE_{A} - EPE_{B} = + 8.3\times 10^{- 4}\ J$

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 = $\frac{\Delta U}{d}$

F = $\frac{8.3\times 10^{- 4}}{0.14} = 5.92\times 10^{- 3}\ N$

(b) The magnitude of the electric field:

$F = qE$

$E = \frac{F}{q} = \frac{5.92\times 10^{- 3}}{1.0\times 10^{- 6}} = 5920\ N/C$