Electric Field and Electric Potential Question

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  • #1
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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, Im 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(Vb-Va)




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(Vb-Va) I already have Vb from part B so I need to find Va

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

Work = Q(Vb-Va)
Work = +1microC ((3.34x105 Volts) -(2.25x105 Volts)) = 0.077Joules

I'm I doing this correctly, I dont have the answers to check my work....

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

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
51
0
Anyone???????
 
  • #3
phyzguy
Science Advisor
4,935
1,872
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.
 
  • #4
51
0
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!
 

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