What is the electric potential at the center of a charged half-circular washer?

AI Thread Summary
The discussion focuses on calculating the electric potential at the center of a charged half-circular washer with a uniform charge Q. The initial approach using the electric field leads to confusion, as the participant incorrectly treats the electric field as constant and does not account for vector addition. It is suggested that starting with the potential rather than the electric field simplifies the problem, as potentials can be added directly. The correct method involves integrating the potential contributions from each differential area element in polar coordinates. The conversation emphasizes the importance of understanding integration in polar coordinates to solve the problem accurately.
annastewert
Messages
12
Reaction score
0

Homework Statement


A plastic circular washer is cut in half and has a charge Q spread uniformly over it. If the electrical potential at infinity is taken to be zero, what is the electric potential at the point P, the center of the old washer? The inner radius of the washer is a, the outer radius is b.
*see attached picture*

Homework Equations


I know that you can use ∫∫E⋅dA=Q/ε and solve for E. Then V(r)=∫E(r)dr

The Attempt at a Solution


∫∫E⋅dA=Q/ε
E∫∫dA=Q/ε
∫∫dA=πb^2-πa^2
E(πb^2-πa^2)=Q/ε
E=Q/ε(πb^2-πa^2)

Then I integrated this, but the answer is V=Q/2πε(b+a) which is not what I am getting. Where am I going wrong?
 

Attachments

  • IMG_5052.JPG
    IMG_5052.JPG
    21.6 KB · Views: 624
Physics news on Phys.org
Your algebra is wrong for two reasons. First, the field at P due to element dA depends on the position of the element dA in relation to P. You took it outside the integral as though it is constant. Secondly, E is a vector, not a scalar. An integral is a sum, and vectors add differently to the way scalars add. The field at P due to different elements will point in different directions, so when you add them there will be cancellation.

The easiest way to solve this problem is to start with potentials, not fields. Potentials also add. What is the potential at P due to an element dA at location (r,θ) in polar coordinates centred on P? What is dA equal to in terms of dr and dθ?
 
I'm confused on how to do this just with potentials, do you use the equation V=Q/(4πεr) and then integrate from a to b? but you would also need to integrate from 0-π wouldn't you? How would you do this?
 
annastewert said:
I'm confused on how to do this just with potentials, do you use the equation V=Q/(4πεr) and then integrate from a to b? but you would also need to integrate from 0-π wouldn't you? How would you do this?
Do you know how to integrate over an area in polar coordinates?
 
I don't think so no.
 
annastewert said:
I don't think so no.
Given a function f(r,θ), the integral over an area A is ∫Af r dr dθ.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top