Recent content by Erik P

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

\begin{equation} 0 = y^2 + z^2 + \frac{2000a^2q_0}{3W_0}-\frac{2}{3} \Rightarrow 0 = y^2 + z^2 - \frac{1}{3} \end{equation} Then r should be sqrt(1/3), right? with a center in (0,0)

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

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

\begin{equation} V(1,y,z) = A(2 - 3y^2 - 3z^2) \end{equation}

Five, I guess?

zero? Sorry, not entirely following here.

I'm not entirely sure to be honest.

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...

I see the problem :D its from 0 to infinity, not 0 to t. Also the t isn't multiplied on the fraction, its only present as an exponent, so t=0 doesn't make the entire thing 0 it just makes the e part equal to 1 and for t = infinity we get e expression becomes 0, which means we have 0 - (above...

I ran it through wolfram alpha to check, and surely you were right about the exponent. I am however a little confused now because I get the following, keep in mind the i had to change the name of some things, like R' is now R_0, V_0 is now v etc. as you can see I now have a negative value and...

Intergrating over the power from 0 to t, at which point I get the same equation :) well sort of :( The problem is the integration of e^(x^(2)), I looked it up and it seems to involve 1/2 sqrt(pi) erf which isn't something I recall having had about in calculus. The answer sheet gives the equation...

Hey, thanks for replying. Was away when you sent that. The instantaneous power dissipated in the resistor is the product of the square of the current and the resistance over which the current runs? \begin{equation} P = I^2R \Rightarrow P' = i'(t)^2R' \end{equation} However this doesn't...

I updated the question so all the details are there.

Hmm... I think my explanation is causing more confusion than anything. Give me a few minutes to write the full original text in.