Question

A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 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 = +9.0 10-4 J.
(a) Find the magnitude and direction of the electric force that acts on the particle. _______N the direction of motion (along, against, perpendicular)
(b) Find the magnitude and direction of the electric field that the particle experiences. _______N/C the direction of motion (along, against, perpendicular)

Answer 1

Answer:

Part a)

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

$E = 4000 N/C$

direction of electric field is along the direction of motion

Explanation:

Part a)

As we know that the change in electric potential energy is equal to the work done by electric field

$W = EPE_A - EPE_B$

$W = 9.0 \times 10^{-4} J$

now from the equation of work done we know that

$W = F.d$

$(9.0 \times 10^{-4}) = F(0.15)$

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

As we know the relation between electrostatic force and electric field given as

$F = qE$

$(6 \times 10^{-3}) = 1.5 \times 10^{-6} E$

$E = 4000 N/C$

direction of electric field is along the direction of motion