U=0.5QV proof. taking the integral

[blackmi4]

I don't feel comfortable with the phrase "take the integral" when it comes to explanations of this equation (energy stored in a capacitor). I've been looking for an explanation of the 0.5 portion of the formula but haven't found one.

Here is the sense I can make of it:

Potential energy is QV. U=QV

For capacitors:
We have a graph with Q on one axis and V on the other.

For some reason we have a diagonal slope with an area underneath it. When we "integrate" we are making an infinite amount of tiny rectangles and finding their areas by multiplying base times height (Q times V). We are summing those rectangles.

For some reason we have a slope which creates a triangular shape underneath it. Due to that triangular shape of the area we are dividing by two to get the above formula with the 0.5 portion in it.

Can anyone explain the formula to me, in particular the integration part?

Thank you!

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[Heday]

I don't feel comfortable with the phrase "take the integral" when it comes to explanations of this equation (energy stored in a capacitor). I've been looking for an explanation of the 0.5 portion of the formula but haven't found one.

Here is the sense I can make of it:

Potential energy is QV. U=QV

For capacitors:
We have a graph with Q on one axis and V on the other.

For some reason we have a diagonal slope with an area underneath it. When we "integrate" we are making an infinite amount of tiny rectangles and finding their areas by multiplying base times height (Q times V). We are summing those rectangles.

For some reason we have a slope which creates a triangular shape underneath it. Due to that triangular shape of the area we are dividing by two to get the above formula with the 0.5 portion in it.

Can anyone explain the formula to me, in particular the integration part?

Thank you!

So where does the 1/2 come from?

ΔPE= qV(average)
So we know that the voltage starts out at 0 and ends at V after the capacitor is fully charged. The average V of the capacitor would be (V+0)/2 or (1/2)V. So we get the following equation:
ΔPE= 0.5qV(max)
We also know that C = (q/v) or q=CV, and when we replace q in the ΔPE equation we get:
ΔPE= 0.5CV^2

Now as far as the integration goes, let's talk about the graph. We have a diagonal slope because when you graph q (x-axis) vs. V (y-axis), your slope is equal to (1/C). Think about it in terms of C = (q/V). Slope is (rise/run). Since our V is on the y axis(rise), q is our run. The plot would then show (V/q) which is equal to (1/C). Diagonal makes sense now? So integration is a fancy way of saying "area under the curve". Our curve is a diagonal line...

We want the area under the red line for a given q, right? So if you draw a line straight up from q to the red line, and then a horizontal line to the y-axis, you have a rectangle with side lengths of q and V, right? Well if you multiply the two sides, you get the area of the rectangle, BUT we want the area under the curve. Notice how that's only half the rectangle? Therefore we take HALF of the q * V. We essentially want the area of the triangle formed by the red line, the x-axis, and a vertical line (imaginary) for which ever value of q we pick. Area of a right triangle is 0.5b*h which is what we are doing here, our base is q, and our height is V... therefore 0.5*q*V.

Hope that helps.

 

[blackmi4]

I understand it now.

Thank you!

 

[gettheleadout]

This is why teaching physics without calculus is a bad idea.

 

[Heday]

This is why teaching physics without calculus is a bad idea.

👍

 

[FeinMS]

dPE=dW=Vdq, Q=CV -> V=Q/C
dW=Qdq/C
PE=W=0.5*Q^2/C = 0.5QV