The differences in electric potential or voltage (we use both terms interchangea

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 V, the change in electric potential energy of that object is given by EPE = q V. When answering the following questions remember that a gain in electric potential energy is given by EPE > 0 and a loss of electric potential energy is given by EPE < 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?

Explanation / Answer

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

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

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

D) U = qV = (-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

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