Electric potential is expressed as:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{{\mathbf{k}}_{\mathbf{e}}\mathbf{q}}{\mathbf{r}}}$, where k_{e} is Coulomb's constant, q is a charge, and r is the distance from the point of the potential to the charge.

**A.**

Let's take point i to be at distance, r = d, and point f to be at distance, r = 3d from the fixed charge.

Initial electric potential, ${\mathit{V}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{k}}_{\mathbf{e}}\mathbf{q}}{\mathbf{d}}$

Final electric potential, ${\mathit{V}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{k}}_{\mathbf{e}}\mathbf{q}}{\mathbf{3}\mathbf{d}}$

A. Suppose we have a positive charge fixed in place. What net change in electric potential will we see if we move from the point I to point f, positive, negative or zero? (Just the net difference; you don’t have to describe what happens during the move). Explain your answer using V = k_{eq}/r

B. What is the electric potential energy of the following pair of charges?

q_{A }= 4.00 x 10^{-7} C

q_{B} = 2.00 x 10^{-6} C

r_{AB} = 3.00 x 10^{-5} m

k_{e} = 8.99x10^{9} Nm^{2}/C^{2}

C. If we were to place a negative charge in between the two positive charges in part B would the change in the total electric potential energy be positive, negative or zero? Explain using PE_{E} = k_{e}q_{1}q2/r_{12}

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