- #1
Erik P
- 33
- 0
Homework Statement
In a specific area of the space, an electrical potential is given as:
\begin{equation}
V(x,y,z) = A(2x^2 - 3y^2 - 3z^2)
\end{equation}
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.
Homework Equations
The radius of a circle displaced from origin to a point (h,k):
\begin{equation}
R^2 = (y - h)^2 + (z - k)^2
\end{equation}
General expression for a circle:
\begin{equation}
y^2 + z^2 +Cy + Dz + E = 0
\end{equation}
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):
\begin{equation}
V = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}
The Attempt at a Solution
Solved the first part a.)
\begin{equation}
\vec{E} = 2A(-2x\vec{i} + 3y\vec{j} + 3z\vec{k})
\end{equation}
\begin{equation}
W_0 = 2q_0Aa^2
\end{equation}
Part b.) I'm not sure how to tackle, here is what I tried...
\begin{equation}
V = A(2x^2 - 3y^2 - 3z^2) = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}
\begin{equation}
r = \frac{q_0}{A(2x^2 - 3y^2 - 3z^2)4\pi\epsilon_0}
\end{equation}
\begin{equation}
A = \frac{W_0}{2a^2q_0}
\end{equation}
\begin{equation}
r = \frac{2a^2q_0^2}{4W_0A(2x^2 - 3y^2 - 3z^2)\pi\epsilon_0}
\end{equation}
Inserting numbers:
\begin{equation}
r = 4.49
\end{equation}
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.
