Electric Potential (I think I'm close?)

Radanie
Messages
1
Reaction score
0

Homework Statement


A -3.2 { nC} charge is on the x-axis at x_1 = -9 {cm} and a 14.6 {nC} charge is on the x-axis at x_2 = 16 {cm}.

***At what point or points on the y-axis is the electric potential zero?


Homework Equations


(1/4*pi*epsilon_0)*[(q1/r2) + (q2/r2)] = 0V


The Attempt at a Solution



The way I've tried solving it is by setting (q1/r2) = (q2/r2)
So 3.2*sqrt((16 cm)^2 + y^2) = 14.6*sqrt((-9 cm)^2 + y^2)
Eventually getting -14644.52 = 202.92 y^2
And y = +/- 8.5

Unfortunately this answer isn't being marked as correct. Does anyone see something wrong with the way I'm trying to solve this?
 
Physics news on Phys.org
What happens when you square -9?

Hint: (y-y_0)^2
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top