Integrating to find the electric field

[SchruteBucks]

Homework Statement

The picture attached shows an insulated board (12m x 4m) with uniform charge density σ. Integrate to find the electric field 8 cm above the center of the board.

Homework Equations

I found the equations $\vec{E}=$$\int\frac{kdq}{r^{2}}\hat{r}$ and dq=σdy (both from google)

The Attempt at a Solution

We've never integrated to find anything in this class before so forgive my crude attempt...
I ended up with $k\int^{10}_{6}\frac{σdy}{y^{2}}$(-$\hat{j}$)

I didn't put anything relating to the width of the board because I'm assuming its part of the density, with σ=Q/V=Q/48m²

Any help would be VERY much appreciated

 

[Spinnor]

Don't you need a double integral? Also won't dq = σdxdy?

 

[SchruteBucks]

Sorry, I should've mentioned that my prof. says that the "symmetry of the situation means only one is needed." As far as if dq=σdxdy, I'm assuming it's just σdy since it's only supposed to be a single integral, but in all honesty, I have no idea. Maybe the other integral isn't needed since we already know the area of the board...?

This is my first time setting up an integral for an electric field, and the only equation I have is the one above, though looking at it now, it looks like it's an equation for a one-dimensional object.

I think I'm more confused now.

 

[Spinnor]

If you knew the electric field due to a line of charge then you could turn this into a one dimensional integral. σ is charge per area so you need to multiply by an area, dxdy, to get charge. I might be missing something here, hopefully others will reply.

 

[asteriatic]

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I would first like to commend you for taking the initiative to search for relevant equations and attempting to use them to solve the problem. It's always important to try and understand the concepts and equations before seeking help.

Now, let's break down the problem and see if we can come up with a solution together. We have an insulated board with uniform charge density σ, and we are trying to find the electric field at a point 8 cm above the center of the board.

First, let's define our coordinate system. Let's say the center of the board is at the origin, and the point we are interested in is at (0, 0.08) m. We can use cylindrical coordinates, with the z-axis pointing upwards, to simplify the integration process.

Next, let's consider a small element of charge dq at a distance y from the origin. The electric field at our point of interest (0, 0.08) will be a vector sum of all these small elements. Using the equation you found, we can write the electric field contribution from this small element as dE = k(dq/r^2) * (-sinθ * r̂), where r is the distance between the element and our point of interest, and θ is the angle between the element and our point of interest.

Now, we need to express dq and r in terms of our chosen coordinates. We know that dq = σdy, where dy is an infinitesimal element of length along the y-axis. We can express r in terms of y as r = √(y^2 + 0.08^2).

Substituting these into our equation for dE and integrating over the entire length of the board (from -6 m to 6 m), we get the final expression for the electric field at our point of interest as E = -2ksinθ * ∫(σy / (y^2 + 0.08^2)) dy.

Now, we just need to evaluate this integral to get our final answer. I'll leave that to you, but it's important to note that the sine term will depend on the angle between the element and our point of interest, which will vary as we integrate over the length of the board.

I hope this helps you to understand the problem better and come up with a solution. Remember, as a scientist, it's important