# The Electric Field of a Continuous Distribution of Charge

1. Jun 8, 2010

### seto6

part 2 The Electric Field of a Continuous Distribution of Charge

1. The problem statement, all variables and given/known data
find the electric potential at point p

2. Relevant equations

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

3. 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.

Last edited: Jun 8, 2010
2. Jun 8, 2010

### 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.

3. Jun 8, 2010

### seto6

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

but does not tell anything about whats 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

4. Jun 8, 2010

### gabbagabbahey

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.