1.

A particle has a charge of +1.5 mu C and moves from point A to point B, a distance of 0.20 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 EPE_(A) - EPE_(B)=+9.0 xx 10^(-4) J. Find the magnitude and direction of the electric force that acts on the particle.

Answer»

`3.0 xx 10^(-3)` N, from A TOWARD B
`4.5 xx 10^(-3)` N, from A toward B
`3.0 xx 10^(-3)` N, from B toward A
`4.5 xx 10^(-3)` N, from B toward A

Answer :B


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