# Electric field at the center of an arc

• tag16
In summary, the electric field at the center of the arc is a function of the opening angle theta. The electric field is highest at the center of the arc and decreases as you move away from the center.
tag16

## Homework Statement

A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the
Figure. What is the electric field at the center of the arc as a function of the opening
angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees.

## Homework Equations

E= KQ/R^2 cos theta

## The Attempt at a Solution

Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look.

where is the figure?

Where'd you get that equation?

Also, if you're worried about how the graph should look, start by figuring it out for some easy points. What's the E field going to be at the center if theta is zero (so all the charge is concentrated in a single point)? How bout if Theta is 360 degrees (so a full circle)? It's a bit harder to figure out for 180 degrees, or 90 or 270 degrees, but you should be able to make a very rough estimate. That will give you some insight into the shape of the graph.

sorry this is the best I can do: http://i848.photobucket.com/albums/ab41/tag16/problem1.jpg?t=1259862072

The equation I found in my physics book except the cos theta part, I was just guessing there. Though it would probably be integral cos theta or sin theta, if it's suppose to be in there at all.

Last edited by a moderator:
$$\lambda$$=Q/$$\pi$$R

dE= kdQ/R^2
dE= (kdQ/R^2) cos$$\theta$$

dQ=$$\lambda$$dl
dl= Rd$$\theta$$
dQ=$$\lambda$$Rd$$\theta$$

dE=(k[$$\lambda$$Rd$$\theta$$]/R^2)cos$$\theta$$

E=$$\int$$(k$$\lambda$$Rcos$$\theta$$/R^2)d$$\theta$$(from $$\pi$$/2 to -$$\pi$$/2)
E=k$$\lambda$$/R$$\int$$ cos$$\theta$$d$$\theta$$
E=k$$\lambda$$/R$$\int$$sin$$\theta$$
E=k$$\lambda$$/R[sin($$\pi$$/2)-sin(-$$\pi$$/2)]
E=k$$\lambda$$/2R
E=k(Q/$$\pi$$R)/2R=2kQ/$$\pi$$R^2
E= 2kq/$$\pi$$R^2

Is this right?

## What is the electric field at the center of an arc?

The electric field at the center of an arc is the force per unit charge experienced by a charged particle placed at the center of the arc. It is a measure of the strength and direction of the electric field at that point.

## How is the electric field at the center of an arc calculated?

The electric field at the center of an arc can be calculated using Coulomb's law, which states that the electric field is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. It can also be calculated using the formula E = (1/4πε0) * (Q/q), where Q is the total charge of the arc and q is the charge at the center.

## What factors affect the electric field at the center of an arc?

The electric field at the center of an arc is affected by the magnitude and distribution of charges on the arc, the distance between the arc and the center, and the dielectric constant of the medium surrounding the arc. It is also affected by any external electric fields that may be present.

## What is the direction of the electric field at the center of an arc?

The direction of the electric field at the center of an arc is determined by the direction of the charges on the arc. If the charges are positive, the electric field will point away from the arc. If the charges are negative, the electric field will point towards the arc. The direction can also be determined using the right-hand rule.

## How does the electric field at the center of an arc affect charged particles?

The electric field at the center of an arc will exert a force on any charged particles placed there. If the charges are free to move, they will experience a net force and accelerate in the direction of the electric field. If the charges are not free to move, they will experience a torque that will cause them to align with the direction of the electric field.

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