Question

The differences in electric potential or voltage (we use both terms interchangeably) can be used to determine changes in the electric potential energy of a charged object. If I have a particle with a charge q that is moving between two points in space which have a difference in electric potential (or voltage) of ΔV, the change in electric potential energy of that object is given by ΔEPE = q ΔV. When answering the following questions remember that a gain in electric potential energy is given by ΔEPE > 0 and a loss of electric potential energy is given by ΔEPE < 0.

A. Suppose a particle with a charge of +7.00 nC moves from a point where the electric potential is 40.0 V to a point where the electric potential is 68.0 V. What is the change in electric potential energy?

B. Suppose a particle with double that charge, +14.0 nC (nanocoulombs) moves from a point where the electric potential is 40.0 V to a point where the electric potential is 68.0 V. What is the change in electric potential energy?

C. Suppose a particle with a charge of +7.00 nC moves from a point where the electric potential is 68.0 V to a point where the electric potential is 40.0 V. What is the change in electric potential energy?

D. Suppose a particle with a charge of -7.00 nC moves from a point where the electric potential is 40.0 V to a point where the electric potential is 68.0 V. What is the change in electric potential energy?

E. Lastly, suppose a particle with a charge of +7.00 nC is released from rest at a point where the electric potential is 68.0 V. Just like a block sliding down a ramp as it loses potential energy, it will gain kinetic energy. What is the kinetic energy of the particle when it has reached a point in space where the electric potential is 40.0 V?

Answer 1

A) ∆U = q∆V = (7nC)×(68-40)V = 7×28 nJ = 196 nJ

B) ∆U = q∆V = (14nC)×(68-40)V = 14×28 nJ = 392 nJ

C) ∆U = q∆V = (7nC)×(40-68)V = 7×(-28) nJ = -196 nJ

D) ∆U = q∆V = (-7nC)×(68-40)V = -7×28 nJ = -196 nJ

E) loss in potential energy = initial potential energy - final potential energy

Loss in potential energy = (7nC)×(68V) - (7nC)×(40V) = 196 nJ

So what loss in electrical potential energy will be that will increase the kinetic energy so kinetic energy of particle will be 196 nJ