Electric potential and capacitance (semicircle problem)

In summary, the problem involves finding the electric potential at the center of a semicircle formed by a uniformly charged insulating rod with a total charge of (-7.50E-6C) and a length of 14cm. Using the formula v= ke \intdq/r, the integral simplifies due to the uniform charge distribution. After solving, the answer is -1.51MV.
  • #1
jimen113
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0

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[tex]\int[/tex]dq/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:
 
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  • #2
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.
 
  • #3
Thanks for your reply.
based on the formula v= ke [tex]\int[/tex]dq/r: I took (1/r) out of the integral and integrated only dq which is (1dq) =Q
 
  • #4
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.
 
  • #5
I used 14cm (0.14m) as the radius and got:
{(8.99x10^9)(-7.50x10^-6)}/(.14m)= -481607.1429
Thank you for your help :)
 
  • #6
Solved!

{(8.99x10^9)(-7.50x10^-6)}/ (.14/pi)= -1513013.461= -1.51MV
 
  • #7
Thank you!
 

1. What is electric potential?

Electric potential is the amount of potential energy per unit charge that a point in an electric field possesses. It is measured in volts (V) and is represented by the symbol V.

2. How is electric potential related to electric field?

Electric potential and electric field are closely related. The electric field is the force per unit charge acting on a test charge, while electric potential is the work done per unit charge in moving a test charge from one point to another in an electric field. They are related by the equation V = -∫E•dl, where V is the electric potential, E is the electric field, and dl is the displacement vector.

3. What is capacitance?

Capacitance is the ability of a system to store electrical charge. It is defined as the ratio of the charge stored on a conductor to the potential difference across the conductor, and is measured in farads (F).

4. How do you calculate capacitance for a semicircle problem?

The capacitance for a semicircle problem can be calculated using the equation C = εA/d, where C is capacitance, ε is the permittivity of free space, A is the area of the semicircle, and d is the distance between the two plates of the capacitor.

5. How does the shape of a capacitor affect its capacitance?

The shape of a capacitor can affect its capacitance because it determines the distance between the two plates. A larger distance between the plates will result in a lower capacitance, while a smaller distance will result in a higher capacitance. Additionally, the shape of the plates can also affect the capacitance. For example, a capacitor with parallel plate geometry will have a higher capacitance compared to a capacitor with cylindrical geometry, even if they have the same distance between the plates.

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