Electric Field and Electric Potential Question

[iurod]

Homework Statement

You have a Square with 10cm lengths on each side. Charges of +4 microC (top right corner), -6 microC (top left corner), and +4 microC (bottom left corner) are located at three corners of the square.

A) Calculate the electric field in the empty corner of the square.
B) Calculate the electric potential at the empty corner of the square?
C) How much work is done if we move a +1 microC charge from the center of the square to the empty corner of the square?

Sorry for not being able to upload the pic, I am kinda not savvy with that kind of stuff.

Homework Equations

Part A:
Electric Field = k(Q/length²)
have to take vector addition into account

Part B:
Electric Potential = k(Q/Length)

Part C:
Electric Potential = k(Q/Length)
Work = Q(V_b-V_a)

The Attempt at a Solution

A)
E due to the two 4 microC charges are the same so:
k(Q/length²)
(9.0 x 10⁹) ((4x10^-6)/0.01)
E = 3.6 x 10⁶ N/C

if I add these two electric field vectorially i get 5.1x10⁶ ([tex]\sqrt{}((3.6x10⁶)²)+(3.6x10⁶)²)[/tex])

Then I find the Electric Field due to the point charge in the top right corner (-6 microC)
k(Q/length²)
(9.0 x 10⁹) ((6x10^-6)/0.0196)
E = 2.76 x 10⁶ N/C

Then since this vector opposes the net of the other two vector I can subtract them to get the overall net Electric Field.

(5.1 x 10⁶ N/C) - (2.76 x 10⁶ N/C) = 2.34x10⁶ N/C

B)
Electric Potential = k(Q/Length)
K((+4microC/0.1) + (+4microC/0.1) + (-6microC/0.14))
Electric potential = 3.34x10⁵ Volts

C)
Work = Q(V_b-V_a) I already have V_b from part B so I need to find V_a

V_a = k((+4microC/0.07) + (+4microC/0.07) + (-6microC/0.07))
V_a = 2.25x10⁵ Volts

Work = Q(V_b-V_a)
Work = +1microC ((3.34x10⁵ Volts) -(2.25x10⁵ Volts)) = 0.077Joules

I'm I doing this correctly, I don't have the answers to check my work...

Thank you very much with your help and time on this.

 

[iurod]

Anyone??

 

[phyzguy]

I think it looks right. The only problem I see is that when you are asked to calculate the electric field, you need to give a vector answer (magnitude and direction), not just magnitude. The rest looks correct.

 

[iurod]

oh, I totally skipped that: it would be Magnitude 5.1x10^6 N/C to the south east (45 degrees below the positive x-axis?

Thank you for your help!

 

[rwisz]

Dear student,

Your calculations for parts A and B seem correct. For part C, your calculation for Va is incorrect. The potential at the center of the square should be the average of the potential at the two 4 microC charges, which would be (k(4x10^-6/0.07) + k(4x10^-6/0.07))/2 = 2.25x10^5 V. Therefore, the work done would be:

Work = Q(Vb-Va)
Work = +1 microC ((3.34x10^5 V) - (2.25x10^5 V)) = 1.09x10^-4 J

Overall, your approach and calculations seem correct. Keep up the good work!