Help with getting an expression for Electric field from a variable Voltage

[quackzilla]

It gives V(x,y,x) = A(x^2,-3y^2,z^2) and asks for E.


What I have done:

E is equal to V/d , and I assume d = (x,y,x)

Dividing them I get E= A(x,-3y,z)

What am I doing wrong?

I also have another question:

In every plane parallel to the xz-plane the equipotential contours are circles. What is the radius of the equipotential contour corresponding to V=1280V and y=2m? (this is an extension of the previous question).

I have a test tomorrow, and there are a few things I still don't know how to do. Any hints on how to do these would greatly help my studying.

 

[Nate810]

For the second question, you need to use the equation E = V/d. The distance d can be calculated using the Pythagorean theorem: d^2 = (x-2)^2 + z^2. Then you can use this to calculate the radius of the equipotential contour.

 

[m2003]

As a scientist, it is important to understand the fundamental concepts and equations in order to solve problems accurately. In this case, the equation for the electric field (E) is not simply equal to V/d. The correct equation is E = -∇V, where ∇ is the gradient operator. This means that the electric field is the negative gradient of the potential, which takes into account the variation of the potential in all three dimensions (x, y, z). Therefore, the expression for E in this case would be E = -A(2x, -6y, 2z).

For the second question, the radius of the equipotential contour can be found by using the equation V = kQ/r, where k is the Coulomb's constant, Q is the charge, and r is the distance. In this case, we can assume that the charge is constant and the potential is given as 1280V. So, the radius can be found by rearranging the equation to r = kQ/V. You can then substitute the given values and solve for r.

As a general tip for studying, it is important to practice solving problems and understanding the underlying concepts. Make sure to also review any relevant equations and their applications. Good luck on your test!