David Griffiths Electrodynamics | Problem 2.2 Solution

[@a.robberts4225]

Your channel may be new, but you have so much potential. This helped me with a university assignment. Thank you!

[@brandonberisford]

I'm honestly so glad to hear this. This channel is sort of a passion project of mine. I wish I could upload at a quicker pace, but life is busy. I'm glad the videos that I have been able to upload have already helped others :) Best of luck with your degree!

[@emmet3690]

great video man, spent a good 2 hours looking for a way to do a problem like this before I found your video. Very helpful with nice hand writing to top it off, thanks!

[@rochehoniball]

Your channel is exactly what I need right now. I'm working through this book this semester and I just know your videos are going to help me stay on track. Thank you for the excellent explanations!! 🌟

[@Nick_Adkins]

Ayy man this is great! I’m working through Griffith’s right now, currently on 2.5 Conductors, and I keep coming back to this problem because I really feel like this problem is the key to understanding how to work out continuous charge distributions. Good stuff! I also do math(and later on physic ) stuff on my channel. Respect 🤙

[@brandonberisford]

Thank you man! I'll check your channel out. Glad the video is helpful to you.

[@carlosaugustooliveira3491]

Parabéns pelas resoluções vc tem me ajudado nas minhas atividades.

[@purdinok229]

Good job with the vid!

[@janoycresva276]

The positive charge would have an electric field E+ pushing away from it with vector components pointing up & right in the z & d axes respectively. The negative charge would have an electric field E- pulling towards it whose vector components would point down & right in the z & d axes respectively. Thus you'd have a sum of d components & sum of z components where for the sum of d components, the equation would be E+sin(theta)+E-sin(theta) which would be kq*0.5d/r^3-kq*0.5d/r^3 because the other charge is negative so the x components would cancel not the z components. For the z components, it would be E+cos(theta)-E-cos(theta) which is kqz/r^3-(-kqz/r^3) & since a negative minus a negative is a positive, the sum of the z components is 2kqz/r^3. Don't rely on your intuition to solve physics problems because there's nothing intuitive about physics whatsoever because intuition would tell us that heavier objects fall faster than lighter ones but Newton's equations predict & show that all objects on earth fall at the same rate regardless of mass in the absence of air resistance. Instead, rely on the rules of mathematics & in particular the rules of vector algebra instead which are objective & won't result in error based on false intuition or bias.

[@jenm1]

But you said it yourself, it's +z +x for q and -z and +x for -q. z cancels, you're left with 2d. You need to do the magnitudes and unit vectors separately, which I agree isn't intuitive at all. When I did this with -d/2 and +d/2, I was getting -2E as in 2E (-xhat) and this doesn't line up with the behavior of charges' fields since the negative charge is in the +x direction.

[@djpanda6853]

Im quite confused on the expression sin theta = Ex/E. Why is not Ex/E1 as in opposite/hypotenuse. Can you explain pls?

[@brandonberisford]

It's because of the angle I chose to be theta. It is the top angle of the triangle, not the bottom.

[@Notra_bel]

@@brandonberisford yeah so? still it should be Ex/E1 ?

[@osamaasif2368]

Thankkkk youuuu so muchhh😊

[@brandonberisford]

Always welcome

[@usman.hqu1]

Hi all i understand but their is problem where you fond Ez when you put limits 0 to L the formula you found should be feee form x but your formula have x why i couldn't understand

[@brandonberisford]

I'm not sure what your question is, I'm sorry.

[@carlosaugustooliveira3491]

Muito bom estou resolvendo algumas questões e essa é interessante!

[@rehmatalishigri4046]

thanks alot sir

[@muqadasnadeem5980]

With x component why did u put sine of theta?

[@brandonberisford]

Look at the triangle I used and where the angle theta is defined. With theta defined as the top right corner, the x component is defined with sin(theta).

[@aroojfatima6555]

sir is'nt sine theta equal to perpendicular/hypot

[@brandonberisford]

It depends how you define your angle theta.

[@liyaphilip7227]

👍

[@zeeshanhaider1131]

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