Electric field problems

In summary, the problem involves finding the electric field at the center of curvature P caused by a uniformly distributed positive charge Q along a semicircle of radius a. The correct method involves finding the field for a segment of the semi-circle and integrating, taking advantage of symmetry. The final answer is (2Qk)/(pi a^2) directed downwards.
  • #1
eku_girl83
89
0
Here an electric field problems I'm struggling with:
Positive charge Q is uniformly distributed around a semicircle of radius a. Find the electric field at the center of curvature P. Answer in terms of Q, k, and a.
When I worked this out on my own, I calculated the magnitude of the electric field to be (2Qk)/(a^2), but this is wrong. Could someone give me a hint on how to do this correctly?
Thanks!
 
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  • #2
Find the field for a segment (dΘ) of the semi-circle and integrate. Luckily it's an easy integral to do (since the distance is the same for all points on the curve). Try and set it up.
 
  • #3
Use symmetry
Consider an axis passing through the Centre and dividing the semicircle into two halves So charge density(lambda)=Q/(pi)r.

Now consider dq charge symmetrical to axis u will see one component of Field produced cancels and other adds up in one direction. Can u show what u get from above hint

Anyway it is going to be moved to HW
 
Last edited:
  • #4
still not getting it :)

Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?
 
  • #5


Originally posted by eku_girl83
Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?
[tex]dQ = \frac {Q}{\pi a} a d\Theta = \frac{Q d\Theta}{\pi}[/tex]
[tex]dE = \frac {k}{a^2} \sin \Theta dQ = \frac{Qk}{\pi a^2} \sin \Theta d\Theta [/tex]


I did neglect to mention, as himanshu points out, that you will need to take advantage of symmetry. Imagine the semicircle intersecting points (-a,0) (0,a) and (a,0). The only component of field you need to worry about is the y-component: by symmetry, the x-components (from opposite sides) cancel.
 
  • #6
answer

I get (Qk)/(Pi a^2) directed down?
Thank you so much for helping me!
 
  • #7


Originally posted by eku_girl83
I get (Qk)/(Pi a^2) directed down?
Thank you so much for helping me!
Check your work. I think you are off by a factor of 2. (Unless I made a mistake.) Did you integrate over the full range of Θ?
 
  • #8
Yes u are off by a factor 2
 

1. What is an electric field?

An electric field is a force field that surrounds an electrically charged object and exerts a force on any other charged objects within the field. This force can either attract or repel other charged objects.

2. How is the strength of an electric field measured?

The strength of an electric field is measured in units of Newtons per Coulomb (N/C). This unit represents the force exerted on a charge of one Coulomb placed in the field. The larger the value of N/C, the stronger the electric field.

3. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude and distance of the charges creating the field. The larger the magnitude of the charges and the closer they are, the stronger the electric field will be. The electric field is also affected by the medium through which it passes, with some materials having a higher or lower permittivity, which can influence the strength of the field.

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction in which a positive test charge would move if placed in the field. The direction is always perpendicular to the equipotential lines, which are imaginary lines that connect points of equal potential within the field.

5. What are some real-world applications of electric field problems?

Electric field problems are used in various real-world applications, such as determining the strength of electric fields around power lines and electrical equipment, designing circuits for electronic devices, and studying the behavior of charged particles in medical imaging techniques like magnetic resonance imaging (MRI).

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