Help With Intergration Dipole Homework Statement

[soupastupid]

Homework Statement

We know the magnitude of the electric field at a location on the x-axis and at a location on the y axis, if we are far from the dipole.

(a) Find \DeltaV= V_p - V_a along a line perpendicular to the axis of a dipole. Do it two ways: from superposition of V due to the two charges and from the integral of the electric field.

(b) Find \DeltaV = V_c - V_d along the axis of the dipole. Include the correct signs. Do it two ways: from the superposition of V due to the two charges and from the integral of the electric field.

Homework Equations

V_q = (1/4\pi\epsilon_0)(+or-q/r)
k= 1/4\pi\epsilon_0

The Attempt at a Solution

(a)
superposition
\DeltaV= (1/4\pi\epsilon_0)(+q/(d^2+(s^2)/4))^(1/2))-(1/4\pi\epsilon_0)(-q/(d^2+(s^2)/4))^(1/2))

\DeltaV= 0

BUT
i don't kno how to do integral
i think its

integral from p to a : E times dd

how do i do it?

i thinks its...

int from a to b: kq/r^2

and i use formula:
int of 1 / x^2 + a^2 dx = (1/a)tan^-1 (x/a)

but I am not sure how to use it or show the answer


(b) superposition



\DeltaV = V_c - V_d
= (E_c)a - (E_d)b = 2kqsa/(a^3) -2kqsb/(b^3)
= ((sq)/(2\pi\epsilon_0))(1/(a^2)-1(b^2))

but again

i don't kno how to do integration

i haven't got a clue for part B

 

[rl.bhat]

At the location on the y-axis
due to +q charge
E= k*q/r^2
E = -dV/dr
dV = - E*dr . V = - k*q* intg(1/r^2)*dr
Find the integration. Similarly find V due to - q. Then find net V.
Follow the same method to find V along x-axis.

 

[soupastupid]

wats the R for V_a?

wats the R for V_p?

are they different?

i think the R for V_a is ((s/2)^2 + d^2 ) )^ (1/2)

but for V_p...

is it just s/2?

if that's true i don't think i will get 0 for \DeltaV = V_p - V_A