Electric Field and Electric Potential

[katie_lynn16]

Hey there, I am new to this site and was wondering if anyone could help me with a problem I'm stuck on:

The electric potential is given at the four corners of a 1 cm square: V1 = 1.6 V, V2 = 12.4 V, V3 = 1.6 V, V4 = -9.2 V. What is the magnitude of the electric field at the centre of the square?

This seems like it should be easy, but for some reason I just can't get it.


Attempt:

Ok, I know that E=deltaV/r and V=kq/r and E= kq/r^2
I've tried finding the differences in the voltage, V1-v2, v2-v3, v3-v4 and v4-v1. then i could find the electric field at that point. Since the voltages arent the same, then the electric field at the center isn't zero.

I'm just really stuck and think that I'm making it harder than it really is.

I also know that the sum of the potential differences around a closed path are equal to zero, but I am not sure if this would help me. Help?

 

[member 553083]

Solution:Using the equation E = deltaV/r, where r is the distance from the center of the square to each corner, you can calculate the electric field at the center. Since the square has a side length of 1 cm, r = 0.5 cm for each corner. For V1 and V2, the electric field is given by: E1 = (12.4 - 1.6)/0.5 = 11.2 V/cm For V2 and V3, the electric field is given by: E2 = (1.6 - 12.4)/0.5 = -10.8 V/cm For V3 and V4, the electric field is given by: E3 = (-9.2 - 1.6)/0.5 = -7.6 V/cm For V4 and V1, the electric field is given by: E4 = (1.6 - (-9.2))/0.5 = 10.8 V/cm The magnitude of the electric field at the center of the square is then the sum of all of these values, or 11.2 + 10.8 + -7.6 + 10.8 = 25.2 V/cm.