Question

Figure 24-40 shows a thin rod with a uniform charge density of 2.40 μC/m. Evaluate the electric potential (in V) at point P if d = D = L/5.00.
http://i46.tinypic.com/e66lhx.png

Answer 1

darknet link

Answer 2

I'll fix it in a sec.

Answer 3

http://i46.tinypic.com/e66lhx.png

Answer 4

V= kQ/r btw

Answer 5

:)

Answer 6

How do I do r if I don't have L?

Answer 7

answer is in terms of L

Answer 8

any questions?

Answer 9

So it would be (8.99e9)(4/5)/sqrt(x^2+L^(2/25))

Answer 10

x=d=L/5?

Answer 11

\[\frac{ L ^{2} }{ 25 }\]
and were you planning to integrate?

Answer 12

not sure where the 4/5 is coming from.... and where's sigma?

Answer 13

I'd like to help you with this but it's difficult if I don't know what level you're at or what you're understanding/not understanding...

Answer 14

This is just Physics with calc.

Answer 15

ok, what I meant was more your personal understanding, like do you get why there's an integral, do you get the geometry etc

Answer 16

I know that I have to integrate, I just wasn't sure what to do with the Ls and how they cancel.

Answer 17

just leave them, the voltage obviously depends on L. eg. if L = 1mile, it means the point P is over a fifth of a mile from the charge distribution, so obviously in a case like that the Voltage would be pretty much zero.

Answer 18

I was looking at (Landa)K ln(L+(L^2+d^2)^.5/d)

Answer 19

oh lambda sorry. I put sigma. my bad...

Answer 20

I should've used lambda for charge/length

Answer 21

sigma is charge/area

Answer 22

sry

Answer 23

it's okay.

Answer 24

d?

Answer 25

d as in the distance from the point to the charge.