Electric field due to a point charge?

Question

How to find potential at a point in the Electric field due to a point charge?Give answer with explaination.

Answer 1

Electric field is an electric force field produced by a single charge or

E=F/q where

W=work done to move a charge q in field E distance d
W=Fd

Also potential difference V can be expressed as

V=W/q=Fd/q=Ed

Thus by measuring potential difference between two points and the distance between these points we can compute the electric field.

E=V/d

Answer 2

What do you mean by THE electric field. Are you saying there are in place charged particles creating a potential field, and you want an answer to what would be the potential at a particular point?

Find me a good electromagnetics book and I'll read it. Mine is dismal. Maybe H. M. Schey or Malvino could write one.

Still, though more time reading his answer is required, I'd back Benjamin N on this result:

Integral(R to infinity)[-kQ/r^2dr]=kQ/R

Answer 3

The electric potential is the potential energy of a test charge of "1" in the field of the point charge. To figure this out, find out how much work you'd have to do done on the charge to bring it from infinitely far away to point r. If your point charg is Q and your test charge is 1, then the force on the test charge is kQ/r^2 in the radially outward direction. (In SI units; in Gaussian units k=1).

Then the work=integral (from infinity to R)[-kQ/r^2 dr] where the angle in the dot product is zero, and the negative sign because the force you have to exert on the the charge is in the opposite direction from the force of the field on the test charge--
Integral(R to infinity)[-kQ/r^2dr]=kQ/R

Answer 4

keep in mind the gap formulation: d = ?((x1 - x2)^2 + (y2 - y1)) r1^2 = (4 - -a million)^2 + (-2 - 0)^2 = 25 + 4 = 29 r2^2 = (a million - -a million)^2 + (2 - 0)^2 = 4 + 4 = 8 E1 = -7k / (r1^2) = -7k/29 E2 = 12k / (r2^2) = 12k/8 = 3k/4 E = 3k/4 - 7k/29 F = qE = a million.67E-19E for area b it is the way you come across the magnitudes. The guidelines are seen and that i can't draw at here. stable luck!

Answer 5

v=1/4
derive by integrating dw = -fdx
-ve sign is becoz work done is opp. to the force experienced

Answer 6

Look in the very beginning of Clerk-Maxwell's "Treatise on Electro-Magnetism"