Field due to ring of charge

In summary, a uniform circular ring of charge Q=4.20 microCoulombs and radius R=1.30 cm is located in the x-y plane, centered on the origin. The magnitude of the electric field E at point P, located at z=3.70 cm, can be calculated using the equation E_z = (kQz)/((z^2+R^2)^(3/2)), where k=9x10^9, Q = 4.2*10^-6, x= 0.0130, and z = 0.0370. However, the correct equation is E_z = (kQz)/((z^2+R^2)^(
  • #1
J6204
56
2

Homework Statement


A uniform circular ring of charge Q=4.20 microCoulombs and radius R=1.30 cm is located in the x-y plane, centered on the origin as shown in the figure.

What is the magnitude of the electric field E at point P, located at z=3.70 cm?

I have worked out the problem and got the following formula which I know is correct but when I enter in my numbers it doesn't seem to give me the correct answer, so I'm wondering what I am doing wrong (see below)

E_z = (kQx)/((x^2+R^2)^(3/2))

where k=9x10^9, Q = 4.2*10^-6, x= 0.0130, and R = 0.0370.

Homework Equations

The Attempt at a Solution


I have worked out the problem and got the following formula which I know is correct but when I enter in my numbers it doesn't seem to give me the correct answer, so I'm wondering what I am doing wrong (see below)

E_z = (kQx)/((x^2+R^2)^(3/2))

where k=9x10^9, Q = 4.2*10^-6, x= 0.0130, and z = 0.0370.

So what am I doing wrong? are my numbers above correct?
****.jpg
****.jpg


2.316×10^7 N/C
 
Last edited:
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  • #2
Looks like you swapped constants. This could be avoided if you restated the equation as

E_z = (kQz)/((z^2+R^2)^(3/2))

R = .0130 m
z = .0370 m
...
 
  • #3
rcgldr said:
Looks like you swapped constants. This could be avoided if you restated the equation as

E_z = (kQz)/((z^2+R^2)^(3/2))

R = .0130 m
z = .0370 m
...
thanks that fixed my problem!(
 

What is the formula for calculating the electric field due to a ring of charge?

The electric field due to a ring of charge can be calculated using the formula E = kQx / (x2 + R2)3/2, where E is the electric field, k is the Coulomb's constant, Q is the total charge of the ring, x is the distance from the center of the ring, and R is the radius of the ring.

What is the direction of the electric field due to a ring of charge?

The electric field due to a ring of charge is always directed perpendicular to the plane of the ring, pointing either towards or away from the center of the ring depending on the location of the point where the field is being measured.

How does the electric field due to a ring of charge change as the distance from the ring increases?

The electric field due to a ring of charge decreases as the distance from the ring increases, following an inverse square law. This means that the field strength decreases by a factor of four as the distance doubles.

What is the difference between the electric field due to a ring of charge and a point charge?

The electric field due to a ring of charge is dependent on the distance from the ring, while the electric field due to a point charge is dependent on the distance from the point. Additionally, the electric field due to a ring of charge is always perpendicular to the plane of the ring, while the electric field due to a point charge can have any direction.

Can the electric field due to a ring of charge ever be zero?

Yes, the electric field due to a ring of charge can be zero at certain points along the axis of the ring, as well as at the center of the ring. This means that there is no net force on a charged particle placed at these points.

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