- #1

- 1,233

- 401

## Homework Statement

- Two particles each with a charge of +3.00 μC are located on the x axis, with one particle at x = -0.80 m, and the other particle at x = +0.80 m.

- a) Determine the electric potential on the y axis at the point y = 0.60 m.

- b) What is the change in electric potential energy of the system if a third particle of charge

q = - 3.00 μC is brought from infinity to the point on the y axis where y = 0.60 m?

- a) Determine the electric potential on the y axis at the point y = 0.60 m.

## Homework Equations

Vp=KQ/r

Vb-Va

Pythagoras Theorem

## The Attempt at a Solution

First I draw the diagram. I place one charge at x=-0.80m and the second at x=0.80m. The point at which I want to calculate the potential energy is at (0,0.60m). I apply the Theorem of Pythagoras to get the radial distance from where the point charge is located to the point I want to find potential.

r=Sqrt(x^2 +y^2)

r=1m

Since I have two charges, the Total potential is equal to the sum of both of the charges.

Vtotal= V1 +V2.

Since Both charges are equal and are the same distance apart to the point p,

Vtotal=53940 V.

This is my answer for part A.

for part B, I want to find the potential difference Vb-Va. A charge comes into point P from infinity, So we say that at infinity it has zero potential. However, at P, the potential is -44950 V.

Now, my problem is. Can I use my result from part a to answer part b. Meaning Va=53940 V.

Since the charges are not moving from P (the ones that moved from the x-axis).

Would my answer be Vb-Va=-44950-53940

Vb-Va=-98,890?

Sorry for this beginner problem. I am not sure if I am understanding the concept of Potential due to Point Charges.