# Problem: 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 = keq/rB. What is the electric potential energy of the following pair of charges?qA = 4.00 x 10-7 CqB = 2.00 x 10-6 CrAB = 3.00 x 10-5 mke = 8.99x109 Nm2/C2C. 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 PEE = keq1q2/r12

###### FREE Expert Solution

Electric potential is expressed as:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{{\mathbf{k}}_{\mathbf{e}}\mathbf{q}}{\mathbf{r}}}$, where ke 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, ${\mathbit{V}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{k}}_{\mathbf{e}}\mathbf{q}}{\mathbf{d}}$

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

83% (409 ratings) ###### Problem Details

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 = keq/r

B. What is the electric potential energy of the following pair of charges?
q= 4.00 x 10-7 C
qB = 2.00 x 10-6 C
rAB = 3.00 x 10-5 m
ke = 8.99x109 Nm2/C2

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 PEE = keq1q2/r12