Question about the magnitude and direction of an electric field.

[peaceinmideas]

Homework Statement

The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 1920x^2, where x (in meters) is the perpendicular distance from plate 1. What is the magnitude and direction of the electric field at x = 2.5 cm? (Take the direction perpendicular to and away from plate 1 to be positive.)

Homework Equations

E= V/r

The Attempt at a Solution

First, I converted the cm to m, and then I plugged everything in, by saying: (1920*0.025^2)/0.025= 48.0. The answer is incorrect, also the negative of 48 is incorrect. I am not sure where to go from here and there was no picture to assist. Any help would be appreciated. Thanks.

 

[rl.bhat]

The relevant equation given by you is true for uniform field. In the given problem field is not uniform. Potential is a function of x^2. In that case the eelctric field is given by
E = - dV/dx.

 

[peaceinmideas]

When I put in my numbers using that equation, I get the same thing: E= -dV/dx = - 1920*(.025 ^2)/ .025 = 48V/m. I may be mistaken though... any thoughts?

 

[rl.bhat]

When I put in my numbers using that equation, I get the same thing: E= -dV/dx = - 1920*(.025 ^2)/ .025 = 48V/m. I may be mistaken though... any thoughts?

It is not correct.
E = -dV/dx = - 1920*d/dx(x^2).
What is the derivative of x^2?

 

[peaceinmideas]

Did you mean to say 1920*x/ dx? Then, I would get (1920*.025)/ (2*.025) = 960, does that sound correct?

 

[rl.bhat]

Did you mean to say 1920*x/ dx? Then, I would get (1920*.025)/ (2*.025) = 960, does that sound correct?

No. d/dx(x^2) = 2x.
E = -dV/dx = - 2*1920*x

 

[peaceinmideas]

Oh, okay. Thanks a lot for all of your help. I am not that great at physics.