A particle has a charge of +1.5 µC and moves from point A to point B, a distance

Question

A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.25 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 = +6.50x10^-4 J.

(a) Find the magnitude and direction of the electric force that acts on the particle.

magnitude in N?
direction ? along against perpendicular to

(b) Find the magnitude and direction of the electric field that the particle experiences.

magnitude in N/C ?
direction ? along against perpendicular to

Explanation / Answer

Given that
Charge q = 1.5 µC = 1.5 x 10^-6 C The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +6.50x10^-4 J d =0.25 m --------------------------------------------------- a) Work is   W = F.d

We have work is also equal to the difference of potential energy

      EPEB - EPEA = F.d              F = (EPEB - EPEA )/d                = (+6.50 x10^-4 J)/(0.25m)                 =26x10^-4 N Since the particle is positive then the force is from point A toward point B.

b) The relation between force and electric field is
       F = qE
       E = F/q           = (26x10^-4 N)/( 1.5 x 10^-6 C)            =  1733.3N/C

The field is from A to B.

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