Got stuck Electric Potential due to Dipole

[IHave]

Homework Statement

Calculate the electric potential due to a tiny dipole whose dipole moment is 6.0×10^−30 C*m at a point 3.6×10^−9 m away for the following cases.

a) This point is along the axis of the dipole nearer the positive charge.

b) This point is 45 degrees above the axis but nearer the positive charge.

c) This point is 45 degrees above the axis but nearer the negative charge. Let V = 0 at r = infinity .

Homework Equations

For part a) I used p = q*d, where p is the dipole moment, q is the charge, and d is displacement. Then used (1/(4*pi*E))*(q/d), where E is the permittivity of free space constant, 8.85 x 10^-12

For part b) I got stuck on this part, I know I have to use the given 45 degrees, perhaps to find a new value for d?

Im not as familiar as I would like to be with the subject, any help or explanation welcome.

The Attempt at a Solution

a) V = 4.2 x 10^-3
b) V = ?
c) V = ?

 

[bjd40@hotmail.com]

for part b and c:
1.draw a neat diagram.
2.calculate the distance of the point at which the potential is to be calculated from each charged end of the dipole.
3. assume the charge and the separation of the charges of dipole.
4.relate this with the dipole moment given ( u can replace this in the eqn. u find later).
5.now find the potential at the point due each charge ( remember to take opposite sign of potential for +ve and -ve charges of the dipole).
6 add this to get the result.
** as the dipole is tiny, may be, u will require to neglect square and higher power of l, the separation of charges of dipole or any other approximation of the sort. i hav'nt done the problem physically, so i cannot be certain.