# How to find potential from 2 charges through 4 points?

## Could someone help with this assignment please? The figure shows two positive charges, Q and 2Q. They are creating an electric potential through the points of A, B, C and D. Write down the following; Which point has the highest, second highest, second lowest and lowest potential.

##### 1 Answer
May 9, 2018

${V}_{D} > {V}_{C} > {V}_{B} > {V}_{A}$

#### Explanation:

Since the potential caused by a positive charge falls off with the distance, this problem can be worked out without any explicit calculation.

• Point D has the highest potential, being closest to both the charges
• point C comes next
• among A and B, B has the higher potential, since it is closer to the bigger charge than A

To verify, here are the actual values

• ${V}_{D} = \frac{1}{4 \pi {\epsilon}_{0}} \left(\frac{Q}{\frac{d}{2}} + \frac{2 Q}{\frac{d}{2}}\right) = \frac{1}{4 \pi {\epsilon}_{0}} \frac{6 Q}{d}$
• ${V}_{C} = \frac{1}{4 \pi {\epsilon}_{0}} \left(\frac{Q}{\frac{d}{\sqrt{2}}} + \frac{2 Q}{\frac{d}{\sqrt{2}}}\right) = \frac{1}{4 \pi {\epsilon}_{0}} \frac{3 \sqrt{2} Q}{d}$
• ${V}_{B} = \frac{1}{4 \pi {\epsilon}_{0}} \left(\frac{Q}{\sqrt{2} d} + \frac{2 Q}{d}\right) = \frac{1}{4 \pi {\epsilon}_{0}} \frac{\left(4 + \sqrt{2}\right) Q}{2 d}$
• ${V}_{B} = \frac{1}{4 \pi {\epsilon}_{0}} \left(\frac{Q}{d} + \frac{2 Q}{\sqrt{2} d}\right) = \frac{1}{4 \pi {\epsilon}_{0}} \frac{\left(1 + \sqrt{2}\right) Q}{d}$