The Electric Field of a Continuous Distribution of Charge

[seto6]

part 2 The Electric Field of a Continuous Distribution of Charge

Homework Statement

find the electric potential at point p

Homework Equations

v=Kq/r...v=Er.

The Attempt at a Solution

using this pic above and dQ=(Q/L)(dX)
v=(K)(dQ)/(L-(X_i)+d) then sub dq=dX(Q/L)

we get...
(KQ/(L))(dX/(L+d-(X_i)) then integrating

V=(KQ/L)((-ln(d))+(ln(L+d)))

is this correct? if not could some one tell me where i went wrong.
thanks in advance.

 

[kuruman]

That looks about right except that I would combine the logarithm arguments and write it as

V=\frac{kQ}{L}ln \left(1+\frac{L}{d}\right)

Then you can see more clearly what happens when d is much larger than L.

 

[seto6]

i see what you are doing but as d>>L it goes to zero..

but does not tell anything about what's below.

a) an infinitely long wire with total charge Q
b) an infinitely long wire with total charge Qd/L
c) a point charge of magnitude Q
d) an electric dipole with moment QL

 

[gabbagabbahey]

i see what you are doing but as d>>L it goes to zero..

Not quite, as d\to\infty V will go to zero. For d\gg L you'll want to use the Taylor series expansion of \ln(1+x) with x=\frac{L}{d} to get an idea of how the potential behaves.