for #1.
I put a gaussian surface around the rod and its the same electric field due to a point charge which is
E = Kq/r^2

q = - Q
r = R + d but since R >> d we can assume d is 0. So r = R
Then the answer would be ...
E = k(-Q) / R^2

for #2
I know that the electric potential difference (V) is the derivative of the Electric field or V = integral E dr.
hmmm im stuck on this 1, do i take partial derviatives or something?

for #3
V = integral of E dr from R1 to r, r is the distance from the center to the gaussian surface.
since E is constant its just E integral dr from R1 to r.
then the answer would be ..
V = E (r - R1) where E is lambda / 2*pi*Epsilon*r

for #4
a) C = 4*Pi*k*epsilon*(ab / b - a)
plug in values and u get C.

B.) sigma (charge per area) = q / A where A is the surface area of the inner sphere.
so sigma = q / 4*pi * (1.20)^2


this is what i have done so far, not sure about a few answers though

This post was edited by FamilyGuyViewer on Apr 2 2013 04:40pm

I don't have much time right now, but for question 1:

The green section of the circle cancels perfectly with the red section of the circle. Leaving just the purple section. You can probably treat the purple section as a straight line of length d.

This post was edited by Azrad on Apr 2 2013 04:55pm

theres only 1 small opening at the top, not 2 openings

you misunderstood the point of azrad's diagram.

I probably did so why dont u explain it than

if you have two point charges with the same charge and you measure the E-field at the halfway point between the two charges, you get zero.

I once knew how to do this problem, but then I took a Kobe to the knee.

Yes that I know.

So is something wrong with #1? I'm not really concerned on #1 but what im more concerned on is #2

. 1 sec

This post was edited by qesh on Apr 3 2013 12:10am

for #2 , yes you need to take the partial derivatives of V
E= - grad(V)
E=- (dV/dx,dV/dy)
E= - (10sqrt(x^2+y^2) +10x^2/(sqrt(x^2+y^2) , 10xy/(sqrt(x^2+y^2))
E=-1/(sqrt(x^2+y^2)(20x^2,10xy)
E(3,4)= -1/(sqrt(9+16)(20*9,10*3*4)
E(3,4)=-(1/5)(180,120)
E(3,4)=-12(3,2)