Calculating Potential Difference in a Uniform Electric Field

In summary, the potential difference between yi= -5cm and yf=5cm in the uniform electric field E = ( 20,000{ i } - 50,000 { j } )V/m is -1450000.
  • #1
aliaze1
174
1

Homework Statement



What is the potential difference between yi= -5cm and yf=5cm in the uniform electric field E = ( 20,000{ i } - 50,000 { j } )V/m?

Homework Equations



∆V = V(sf) - V(si) = -∫ Es ds

with the limits of the integral being sf and si

The Attempt at a Solution



I tried doing the integral, resulting in (E2)/2, and then multiplying proceding as so:

E= √ (200002 + 500002) = 53851.64807
(E2)/2 = 14500000 = x
x |0.05,-0.05 = -{x(0.05) - x(-0.05)} = -{2[x(0.05)]} = -1450000 = incorrect

any help?

thanks!
 
Physics news on Phys.org
  • #2
I think you're taking the integral as if it were E dE instead of ds (Compare with the case where if it were the integral of x dx, then the integral is x^2/2). E is a constant since it is "uniform", so you can treat it as a constant and pull it out of the integral. If you're only dealing with one dimension (along y), then the integral of E ds simplifies to E(Sf - Si).
 
  • #3
Anadyne said:
I think you're taking the integral as if it were E dE instead of ds (Compare with the case where if it were the integral of x dx, then the integral is x^2/2). E is a constant since it is "uniform", so you can treat it as a constant and pull it out of the integral. If you're only dealing with one dimension (along y), then the integral of E ds simplifies to E(Sf - Si).

aaahh lol nice...yea that was my mistake...lol whenever i see the integral sign i jump to conclusions without looking at the second part

thanks!
 
  • #4
well actually...i tried that and it didn't work:

E = 53851.64807

Sf-Si = 0.05 - (-0.05) = 0.1

E * (Sf-Si) = 53851.64807 * 0.1 = 5385.164807 = Incorrect :(

Any help?

Thanks!
 
  • #5
I have only one attempt left. I noticed that I didn't put the negative sign there, but the computer would tell me to 'check your signs' if this was a sign issue...so i assume it isn't
 
  • #6
any help?
 

What is Potential Difference?

Potential Difference, also known as Voltage, is the difference in electric potential energy between two points in an electric field.

How is Potential Difference measured?

Potential Difference is measured in units of volts (V) using a voltmeter.

What is the relationship between Potential Difference and Electric Field?

Potential Difference is directly proportional to the strength of the Electric Field. This means that as the Electric Field increases, the Potential Difference also increases.

Can Potential Difference be negative?

Yes, Potential Difference can be negative. This indicates that the direction of the electric current is opposite to the direction of the Electric Field.

How does Potential Difference affect the flow of electric current?

Potential Difference is the driving force for the flow of electric current. A higher Potential Difference will result in a higher current flow, while a lower Potential Difference will result in a lower current flow.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
151
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
308
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
361
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
1
Views
137
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top