I agree. E does not equal E times cosine of theta. However, the dot product of E and dr does equal E cosine theta dr. And yes, it probably would have been good to put the vector symbol over the E in the very first integral equation.
How is the length h a constant value if it is the same as r? This means that you cannot take h out of the integral and you end up using a natural logarithm which leads me to the conclusion that you are trying to use such a mathematical proof in physics but it does not work. On the other hand, if h and r would really be different variables then you could take h out of the integral, and after integrating 1 with respect to r you get r. Then you would need to plug in the values of b and a_2 into r, subtract them and end up with -E * d * r * h. None of these methods seems to work as proof that a potential difference remains the same no matter the path between two potentials. Please clarify this.
FYI: It's helpful to put times in a comment so people know which part of an 8.5 minute video you are referring to. h and r are different. We substitute h into the "cosine theta equals adjacent over hypotenuse" equation because h is the length of the hypotenuse, which is constant.
@@FlippingPhysics I was thinking the same thing(Just 8 month later). I didn't understand what you meant so this might clarify it a bit. But first, (I think Flipping physics made a mistake by saying Cosine theta equals opposite over hypotenuse in the comments, when it's "adjacent" over hypotenuse, but its correctly stated in the video at 1:45 .) Pretty much Cos(Theta) is a constant. Whatever you replace cos(theta) with, here being d/h(d over h) will be a constant regardless of there actual change in size. d/h is a ratio that is constant and not to be considered a single length that may be extended or shortened. So you can have d = 5 and h =10 or d= 10 and h= 20 and their ratio would still be **d/h = 1/2** which when considering their arccos is 60 degrees, or a constant. Not sure if I'm either overthinking it and got it wrong, but that's how I understood it. Please know that that specific ratio or angle is just an example, I don't know the actual angle or ratio. Thanks Flipping Physics, I'm so tight on time and this is helping me a lot.
Correct me if I am wrong, but shouldn't there be a negative sign in the work equation in the bottom left of the video at 3:50?
Music to my ears
Nice
Another point to mention is that E is not equal to E times cosine of theta. The first E must have a vector symbol above it.
I agree. E does not equal E times cosine of theta. However, the dot product of E and dr does equal E cosine theta dr. And yes, it probably would have been good to put the vector symbol over the E in the very first integral equation.
How is the length h a constant value if it is the same as r? This means that you cannot take h out of the integral and you end up using a natural logarithm which leads me to the conclusion that you are trying to use such a mathematical proof in physics but it does not work. On the other hand, if h and r would really be different variables then you could take h out of the integral, and after integrating 1 with respect to r you get r. Then you would need to plug in the values of b and a_2 into r, subtract them and end up with -E * d * r * h. None of these methods seems to work as proof that a potential difference remains the same no matter the path between two potentials. Please clarify this.
FYI: It's helpful to put times in a comment so people know which part of an 8.5 minute video you are referring to.
h and r are different. We substitute h into the "cosine theta equals adjacent over hypotenuse" equation because h is the length of the hypotenuse, which is constant.
@@FlippingPhysics I was thinking the same thing(Just 8 month later). I didn't understand what you meant so this might clarify it a bit. But first, (I think Flipping physics made a mistake by saying Cosine theta equals opposite over hypotenuse in the comments, when it's "adjacent" over hypotenuse, but its correctly stated in the video at 1:45 .) Pretty much Cos(Theta) is a constant. Whatever you replace cos(theta) with, here being d/h(d over h) will be a constant regardless of there actual change in size. d/h is a ratio that is constant and not to be considered a single length that may be extended or shortened. So you can have d = 5 and h =10 or d= 10 and h= 20 and their ratio would still be **d/h = 1/2** which when considering their arccos is 60 degrees, or a constant. Not sure if I'm either overthinking it and got it wrong, but that's how I understood it.
Please know that that specific ratio or angle is just an example, I don't know the actual angle or ratio.
Thanks Flipping Physics, I'm so tight on time and this is helping me a lot.