Homework Help: Show that the equipotential lines are circles

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1. Mar 19, 2016

Erik P

1. The problem statement, all variables and given/known data
In a specific area of the space, an electrical potential is given as:

V(x,y,z) = A(2x^2 - 3y^2 - 3z^2)

where A is a constant.

a.) Determine the electrical field E for any given point in the area. A test charge q_0 is moved from the point (x,y,z)=(a,0,0) to origin (0,0,0). Determine the expression for the work W_0 performed on the test charge by the electrical force. Determine A as a function of W_0, q_0, and a.

b.) Show that the equipotential lines in planes parallel to the (yz)-plane are circles. Determine the radius of the equipotentiallines where x=a=1.0 m and V=1000 volts, for q_0 = 1.0 micro columb and W_0 = 2.0 mJ.

2. Relevant equations
The radius of a circle displaced from origin to a point (h,k):

R^2 = (y - h)^2 + (z - k)^2

General expression for a circle:

y^2 + z^2 +Cy + Dz + E = 0

Voltage due to point charge (really not sure if this is the right one to use, i tried inserting numbers and it gives me the wrong result):

V = \frac{q_0}{4\pi\epsilon_0R}

3. The attempt at a solution

Solved the first part a.)

\vec{E} = 2A(-2x\vec{i} + 3y\vec{j} + 3z\vec{k})

W_0 = 2q_0Aa^2

Part b.) I'm not sure how to tackle, here is what I tried...

V = A(2x^2 - 3y^2 - 3z^2) = \frac{q_0}{4\pi\epsilon_0R}

r = \frac{q_0}{A(2x^2 - 3y^2 - 3z^2)4\pi\epsilon_0}

A = \frac{W_0}{2a^2q_0}

r = \frac{2a^2q_0^2}{4W_0A(2x^2 - 3y^2 - 3z^2)\pi\epsilon_0}

Inserting numbers:

r = 4.49

the answer is supposed to be 0.6 m, so obviously I'm way off. I suspect it has to do, atleast in part with me using the voltage for a point charge. Not sure how to go about this. I also need to show that they are circle, but that's also a little over my head at the moment.

2. Mar 19, 2016

ehild

You are given the potential finction V(x,y,z). It is not the potential of a point charge. You need the curve in a plane parallel to the (y,z) plane. What do you know about x in that plane?

3. Mar 19, 2016

Erik P

I'm not entirely sure to be honest.

4. Mar 19, 2016

ehild

What is x on the (y,z) plane?

5. Mar 19, 2016

Erik P

zero? Sorry, not entirely following here.

6. Mar 19, 2016

ehild

yes.
And what is x on the blue plane, parallel with the (y,z) plane?

7. Mar 19, 2016

Erik P

Five, I guess?

8. Mar 19, 2016

ehild

yes. Read the problem, you are given that x=a=1.0 m and V=1000 volts. How the potential function looks like if x= 1.0 m?

9. Mar 19, 2016

Erik P

V(1,y,z) = A(2 - 3y^2 - 3z^2)

10. Mar 19, 2016

Erik P

oooohhhh.. i think i see it, give me a few mins to look at it

11. Mar 19, 2016

ehild

You want the line where V=1000 V. What is its equation?

12. Mar 19, 2016

Erik P

1000 = A(2 - 3y^2 - 3z^2)

0 = A(2 - 3y^2 - 3z^2) - 1000

A = \frac{W_0}{2a^2q_0}

0 = \frac{W_0}{2a^2q_0}(2 - 3y^2 - 3z^2) - 1000

13. Mar 19, 2016

ehild

Can you bring it to the standard from of a circle? Where is the center? What is the radius? You also know that a=1. ( x=a=1.0 m) Wo and qo are also given.

14. Mar 19, 2016

Erik P

0 = y^2 + z^2 + \frac{2000a^2q_0}{3W_0}-\frac{2}{3} \Rightarrow 0 = y^2 + z^2 - \frac{1}{3}

Then r should be sqrt(1/3), right? with a center in (0,0)

Last edited: Mar 19, 2016
15. Mar 19, 2016

ehild

Yes. Good work!

16. Mar 19, 2016

Erik P

Thank you for the help and also for being patient with me :)

17. Mar 19, 2016

ehild

You are welcome