Find the linear charge density lambda and the radius of a charged semicircle

In summary, the electric field for a semi-circle is equal to -πKλ/2R, with a value of 10 N/C in this case. The electric field for a straight wire is equal to 2Kλ*(1-2y/√(4y^2+L^2)), with a value of 8 N/C in this case. To find the values of R, λ, and the length of the wire, you need a third equation. By substituting L = 2*π*R, you can solve for R = 1.18962, L = 7.4746, and λ = 8.41483E-10.
  • #1
Sonic98
3
0
Homework Statement
A uniformly loaded semicircle-shaped wire of radius R has a linear density of charge lambda. An electric field magnitude of 10 N / C is measured at the center of the semi-circle. We straighten this same wire and we note an electric field magnitude of 8 N / C, this time at a distance of 2 m on its perpendicular bisector; the field thus measured is directed towards the wire.

a) Calculate the values of the lambda and R
b) How many electrons would need to be added or removed from the wire in order for it to become neutral? Specify whether to add or remove electrons
c) Calculate the vector electric force on a proton placed at the center of curvature of the wire
when the latter is in the shape of a semicircle
Relevant Equations
At the center of the semi-circle an electric field magnitude of 10 N / C
Straighten wire has an electric field magnitude of 8 N / C
semi-cercle.png


Electric field for the semi-circle

$$E = - \frac {πKλ} {2R} $$

In this case E is equals to 10 N/C

Electric field for the straighten wire

$$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$

In this case E is equals to 8 N/C

What I'm searching is R, λ, and the length of the wire, so I think I may need a third equation ?

Not sure if I took the right approach for this.

Thank you.
 
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  • #2
what is the relationship between L and R?
 
  • #3
L is equal to 2 * π * R ?
 
  • #4
Sonic98 said:
L is equal to 2 * π * R ?
Right, so substituting that you have two equations and two unknowns, yes?
 
  • #5
That's right, so I can do:

$$E = - \frac {πKλ} {2R} $$

$$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$

$$ L = 2 * π * R $$

Which should give me: R = 1.18962 L = 7.4746 Lambda: 8.41483E-10

Now I can pass to the next questions. Thank you :) !
 

1. What is a linear charge density?

A linear charge density is a measure of the amount of electric charge per unit length along a line. It is denoted by the symbol lambda (λ) and has units of coulombs per meter (C/m).

2. How do you find the linear charge density of a charged semicircle?

To find the linear charge density of a charged semicircle, you need to know the total charge and the length of the semicircle. The linear charge density can be calculated by dividing the total charge by the length of the semicircle.

3. What is the formula for calculating linear charge density?

The formula for calculating linear charge density is λ = Q / L, where λ is the linear charge density, Q is the total charge, and L is the length of the line.

4. How is linear charge density related to electric field?

Linear charge density is directly related to electric field. The electric field at a point on the line is equal to the linear charge density multiplied by the distance from the point to the line. This relationship is described by Coulomb's Law.

5. Can the radius of a charged semicircle affect the linear charge density?

Yes, the radius of a charged semicircle can affect the linear charge density. As the radius increases, the length of the semicircle also increases, resulting in a decrease in the linear charge density. Similarly, a decrease in radius would result in an increase in linear charge density.

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