Electric potential and capacitance (semicircle problem)

[jimen113]

Homework Statement

A uniformly charged insulating rod of length 14cm is bent into the shape of a semicircle (looks like a "C"). The rod has a total charge of (-7.50E-6C). find the electric potential at O, the center of the semicircle

Homework Equations

v= ke\intdq/r

The Attempt at a Solution

after integrating I get: KeQ/r
=Ke(-7.50E-6)/(0.14m)
For the semicircle will the radius be 0.7m? I'm frustrated with this problem and maybe I'm just making a dumb calculation mistake.
The answer is: -1.51MV
Thank you in advance! :shy:

 

[Delphi51]

will the radius be 0.7m?

No. The distance around the semicircle is (pi)*r = 14 cm
It would be interesting to see how you set up your integral.

 

[jimen113]

Thanks for your reply.
based on the formula v= ke \intdq/r: I took (1/r) out of the integral and integrated only dq which is (1dq) =Q

 

[Delphi51]

Oh - of course - you don't have much of a job integrating because every bit of the charge is the same distance from the center! So, have you now got the right answer? If not, let us know what radius you used and how it worked out.

 

[jimen113]

I used 14cm (0.14m) as the radius and got:
{(8.99x10^9)(-7.50x10^-6)}/(.14m)= -481607.1429
Thank you for your help :)

 

[jimen113]

Solved!

{(8.99x10^9)(-7.50x10^-6)}/ (.14/pi)= -1513013.461= -1.51MV

 

[jimen113]

Thank you!