Electricity and Magnetism Question

Davio

1. The problem statement, all variables and given/known data
The points A,B,C are the vertices of an equilateral triangle, of side L. Two equal positive charges q are at A and B.

A)Find the electric potential at C assuming V=0 at infinity
B) Give the expression for the work necessary to bring a positive charge Q from infinity to C
C)Write the expression for the work necessary to assemble the system of 3 charges, consisting of two charges q at A and B and a charge Q at C
Consider the system of three charges. By differentiating the elctric potential,determine the electric field E at the midpoint between the two charges q.

Consider now a system of 2 charges, a point charge q>0 at (x,y,z) = (a, 0 , 0) and a point charge -q/2 at (-a,0,0)Show that the quipontential surface V=0. ie. with the same potential than at infinity is a spherical surface. Determine the centre and the radius of the sphere.

2. Relevant equations



3. The attempt at a solution
OK, I can do all the homework sheets, but these exam questions are killing me :-(, heres my attempts:#

We have a triangle, with 2 positive charges and one unknown charge.
If the unkown is positive we lose energy and if negative we gain energy.

Possible pairs
q1 q2 q2q3 q1q3 (I've divided by half already)
I assume that we use something simliar to [tex]\int E.da[/tex]= Qinternal/Eo

or we use the superposition principle, and add A and B together. using coloumbs law. E=k qi, q2 / r^2

In the first case, I have no idea.
IN the second case, I think r^2 is just length A and length B so:

E=k. q ^2
A.B
Is this horribly wrong as I believe it is?

For part B.
I use phi=qi over 4 pi Eo Rij

along with W=qphi

to figure out.

U= q1 q2 over 4 pi Eo r12

in which case we have

U= q^2 over 4 pi Eo (2 (sqrt(half c^2 +D^2))

Part C
I have no idea, Do we use the previous formula in some manner?

Part D

du/dr = q1q2/ 4pi Eo r^2

From there, I have no idea, would r be half AB?

Part E
Sorry :-(

So yeah, this is quite shameful really, didn't expect not to be able to do any of this paper's B section!

carmen77

I definently could be wrong as I'm taking physics 2 right now and will have to drop it haha but here's my shot at part A.

Since V=kq/r (assuming V=0 at infinity), and potential difference is a scalar quantity, meaning direction doesn't have to be taken into account, then to find the potential difference at point C, we would just take the [tex]\Sigma[/tex](kq/r) of the two point charges that are a distance L, (because triangle is equilateral), from point C.

Therefore the potential difference at point C--V=2kq/L

carmen77

Part B W=[tex]\Delta[/tex]KE=-[tex]\Delta[/tex]U

Vq=U
In part A we found out that the potential difference at point C is V=2kq/L

U=(2kq/L)*Q=2kqQ/L

W=-[tex]\Delta[/tex]U=-2kqQ/L

Davio

Hi, thanks for replying!
As I do not underestand how to do this question.. I'll wait for others to answer too before taking your answer as legit. It seems to make sense for A, B i'm not so sure about as I copied what I wrote in my answer off a book..
Thanks again.

Davio

Bump, anyone?

rl.bhat

We have a triangle, with 2 positive charges and one unknown charge.
There is mention of the third charge at C.
A)Find the potential at C due to charges at A and B.
B)Work done on a charge = Q*potential difference.
C) Energy in a system of charges = k*q1*q2/r. So total PE of the system = k(q^2 + 2qQ)/r