Electric field on the axis of a ring-shaped charged conductor

  • #1
yashboi123
17
0
Homework Statement
A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = 0.130 nC uniformly distributed around it
Relevant Equations
E = F/q
1706663702409.png

Hello. I was wondering why do we not multiply cos(alpha) by 2. I believe we should do this since the y-components of the electric field cancel out, meaning there would be 2 x-components of the electric field(at least I think so). Currently, this derivation/answer only considers one horizontal component, not the other half.
 
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  • #2
How does it only consider one horizontal component when the integration is from ##0## to ##2 \pi##?

The integration bounds take care of this “doubling up effect”.

If you did it from ##0## to ##\pi## you would certainly have to multiply by 2 at the end.
 
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FAQ: Electric field on the axis of a ring-shaped charged conductor

What is the electric field on the axis of a ring-shaped charged conductor?

The electric field on the axis of a ring-shaped charged conductor is directed along the axis of the ring. It can be calculated using the principle of superposition, integrating the contributions of all infinitesimal charge elements on the ring. The field at a point on the axis a distance z from the center of the ring is given by the formula: \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Qz}{(z^2 + R^2)^{3/2}} \), where \( Q \) is the total charge on the ring and \( R \) is the radius of the ring.

How does the distance from the center of the ring affect the electric field on the axis?

The electric field on the axis of the ring decreases as the distance \( z \) from the center of the ring increases. Specifically, the field is inversely proportional to \( (z^2 + R^2)^{3/2} \). At points very far from the ring (where \( z \gg R \)), the field diminishes approximately as \( 1/z^2 \), behaving similarly to the field of a point charge.

What happens to the electric field at the center of the ring?

At the center of the ring (where \( z = 0 \)), the electric field is zero. This is because the contributions from all the charge elements on the ring cancel each other out due to symmetry, resulting in no net electric field at that point.

How does the radius of the ring influence the electric field on its axis?

The radius \( R \) of the ring influences the magnitude of the electric field on the axis. For a fixed distance \( z \) from the center, a larger radius results in a weaker electric field because the charge elements are spread out over a larger circumference, reducing the field's intensity at any given point on the axis.

What is the direction of the electric field on the axis of a ring-shaped charged conductor?

The direction of the electric field on the axis of a ring-shaped charged conductor depends on the sign of the charge on the ring. If the ring is positively charged, the electric field points away from the ring along the axis. If the ring is negatively charged, the electric field points towards the ring along the axis.

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