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Electric Potiential

  • Thread starter minifhncc
  • Start date
  • #1
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Electric Field

Hi, first of all I've just found this forum after a few hours of looking at Wikipedia and other websites for the theory and formulas behind this solution but no avail. Thank you to anyone in advanced who can assist me, it'd be much appreciated. Please note that I am usually a person that likes to find out answers for myself but this question has stumped me unfortunately.

Homework Statement


Two plates are 40mm apart and are charged to the potentials of +300V and -100V with respect to earth as shown in Figure A (see attachment)

a) What is the magnitude and direction of the electric field between the plates?

b) What are the potentials at points A and B (A is 10mm from the bottom plate and B is 30mm from it)

c) Calculate the work done in moving a -4uC charge directly from A to B.

Homework Equations


E=F/q

V=W/q

(There may be others, but as I was unable to solve the problem I don't know)

The Attempt at a Solution


(Could not find any solutions, or anything close to it -- Was not able to find the variables for the equations above)

Thanks again in advanced.
 

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Answers and Replies

  • #2
46
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Hi again,

I think it's pretty clear that I do not know the correct formulas for finding the solution to the question I've posted.

Could someone just please let me know of the formulas that I need to solve this problem, as I'd really like to solve this myself without too much help from other people.

Thanks in advanced.
 
  • #3
alphysicist
Homework Helper
2,238
1
Hi minifhncc,

Here is a hint. We normally assume that the electric field is constant between plates such as these. When the electric field is constant, the relationship between E and the change in potential between two points (delta V) is:

[tex]
\Delta V = - E d \cos\theta
[/tex]

where [itex]\theta[/itex] is the angle between the electric field direction and the direction of the displacement d between the initial and final points.
 
  • #4
46
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Hi thanks for the info, may I ask though, why is there a negative sign in the equation that you've given?

Thanks again for your help.
 
  • #5
alphysicist
Homework Helper
2,238
1
It comes from a definition of the potential energy: the work done by a conservative force is equal to the negative of the change in the potential energy corresponding to that force.

For a constant electric field (and constant force):

[tex]
\begin{align}
W_{\rm electric} &= -(\Delta U_{\rm electric})\nonumber\\
F d \cos\theta &= -(\Delta U)\nonumber\\
q E d\cos\theta &= - q (\Delta V)\nonumber
\end{align}
[/tex]

(U is electric potential energy; V is electric potential.) Cancelling the q's and switching the minus sign gives the result.

One result of this is that if a particle is moving in the same direction as the field then the potential along the path is decreasing.
 
  • #6
46
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Thanks for your info mate. Very appreciated!
 

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