Electric field of a charged arc

[brett812718]

Homework Statement

A charge of 18 nC is uniformly distributed along a straight rod of length 4.7 m that is bent into a circular arc with a radius of 2.4 m. What is the magnitude of the electric field at the center of curvature of the arc?

Homework Equations

E=KQ/R^2

The Attempt at a Solution

dQ=\lambdads
\lambda=Q/\piR
dQ=\lambdaRd\Theta
magnitude dE=kdQ/R^2=K\lambdaRd\Theta/R^2=KQd\Theta/R^2
sin\Thetamagnitude dE=KQd\Thetasin\Theta/R^2
magnitude Ey=KQ[-cos\Theta]from a to b/\piR^2

the limits of integration would be a=0 and b=4.7(2\Pi)/2\PiR
I got E=4.31e-1 n/c but it was wrong.

 

[dynamicsolo]

EDIT: On looking through all of what you wrote, I can tell you this.

It is not correct that \lambda = Q/\pi R , but you can just use

dQ = \lambda R d\theta.

However, in your field calculation, look at what field component cancels out in the integration. (You won't integrate dE sin\theta...).

Make a picture of the arc and the point at its center. Choose an axis through the center that is symmetrically placed through the arc; this will let you take advantage of a symmetry consideration and also clarify how to set up the field integration.

 

[brett812718]

thanks

 

[SamTaylor]

Why is dQ = \lambda R d\theta ?
When \lambda has the Unit [\lambda] = \frac{C}{m}

For a circle I understand it.
dQ = Q \left( \frac{dl}{2 \pi r} \right) = \lambda dl

 

[Gear300]

Make sure to take into account that the electric field at a point is a vector.

 

[SamTaylor]

Don't I do that with dE_x = dE cos(\theta)

E = E_x = \int\limits_{-\theta }^{\theta} dE cos(\theta)

I thought dQ = \lambda R d\theta is the charge density, which is a scalar.

 

[cryptoguy]

Why is dQ = \lambda R d\theta ?
When \lambda has the Unit [\lambda] = \frac{C}{m}

For a circle I understand it.
dQ = Q \left( \frac{dl}{2 \pi r} \right) = \lambda dl

You need to express dQ as a function of d(theta) so you can integrate from 0 to 2*(pi) (if it's a circle)
\lambda = \frac{18nC}{4.7m}

 

[SamTaylor]

You need to express dQ as a function of d(theta) so you can integrate from 0 to 2*(pi) (if it's a circle)
\lambda = \frac{18nC}{4.7m}

I think I did not express myself the right way. Sorry

The only thing I don't understand is why there is a R inside dQ = \lambda R d\theta <br />. The first time i tried to solve it I used dQ = \lambda d\theta
Because as you said, I need a function of theta.
This is how it worked for the circle with dQ = \lambda dl

So for the arc it is \left[ \lambda \right] = \frac{C}{\circ m}
I can't interpret that geometrically, it seems to me just to make it fit right

 

[alphysicist]

Hi SamTaylor,

I think I did not express myself the right way. Sorry

The only thing I don't understand is why there is a R inside dQ = \lambda R d\theta <br />. The first time i tried to solve it I used dQ = \lambda d\theta
Because as you said, I need a function of theta.
This is how it worked for the circle with dQ = \lambda dl

So for the arc it is \left[ \lambda \right] = \frac{C}{\circ m}
I can't interpret that geometrically, it seems to me just to make it fit right

As a first look, we know it can't be dQ = \lambda d\theta because that doesn't have the right units--coulombs on the left, and (coulombs/meter) on the right.

The linear charge density \lambda is the charge per length, and here the length is along the arc. So if s is the length along the arc, a true statement to begin with would be

dQ = \lambda\, ds

If you then think about the relationship between the length of a circular arc s, the radius r, and the angle \theta, you'll get the right formula.

 

[SamTaylor]

Thanks a lot.