David Griffiths Electrodynamics | Problem 2.4 Solution

[@mnakooda489]

thank you for these videos you are helping me greatly , please keep on making them we all appreciate it

[@shuvamghosh1542]

Great sir!! You are superb. 💯 Respect and love from India I really love this. I'm disappointed to see the number of your subscriber. But don't worry you are a real gem I am subscribed and also share this content.

[@spacedee6968]

Diligently explained. Thank you!

[@sairaafzal4863]

Thank you so much sir. You nailed it.

[@mnakooda489]

and we've taken 2 point charges, continuous uniform line charges acting as point charges, so shouldn't it be 2 lambda z over 4pi epsilon

[@swayamsevak7557]

Thank you so much for an awesome explanation. I have a silly question however, probably because I am not able to grasp it intuitively. I am repeatedly using Pythagoras Theorem to equate curly r as √ [z^2 + (a/2)^2] given that we are "moving" the point from the top of linear segment to the top of the square's center. And that is causing me problems. I do understand that if the distance between P(original) and edge of linear element is z , then we are doing nothing except changing the orientation of the point (where it becomes curly r) so that the distance between P(new) and edge of linear element is unchanged at z! Can you please help me here, so that I can naturally and intuitively grasp it😣... (Edit- Dear Brandon, a way to essentially convince myself that it has nothing to do with Pyth Theorem could be that the "rotation" P has made from top of linear segment to top of square's center is not a "straight line distance" but a curved arc, so that it renders any such Pythagorean Theorem application as INVALID. Is this correct?)

[@brandonberisford]

Hey Swayam. Thanks for your question. If I'm understanding it correctly, you're confused why the distance "z" from the original P and the edge is different from the distance between the new P and the edge? The distance between two points right above each other you can think of like the edge of a triangle, and when the point moves to the new spot it then becomes a hypotenuse of a triangle. The distance relationship there is different. I think depending on the value of "a", there is a case where it could be equal, however in the general case this would not be true. You could set z = sqrt(r^2 + (a/2)^2) and solve for a to find what a must be for that relationship to hold true, but in general it is not true hence why were using Pythogora's theorem. If you want to convince yourself, get a long ruler or meter stick and try measuring the distances yourself. draw two points on a piece of paper directly above each other, measure the distance. Then move that point horizontally over some and measure the new distance and you'll find the distances are not equal. However there may be a certain point you move it over where they do equal each other, but its not the case always.

[@swayamsevak7557]

@@brandonberisford @Brandon Berisford Thank you for your prompt response. But shouldn't they be equal if we try to think of the same in coordinates? Essentially when we are moving from P (old) to P(new) only the y (or whatever axis we specify to be horizontal) coordinate is changing. So if P old is some (x1, y1, z1) then P new is just (x1, y2, z1) no? So my question was more along the lines of why they are coming out to be sort of same. And if it is indeed becoming a hypotenuse then why can't we use Pyth Theorem to relate curly r and z and z' (If z is distance from edge to P(old) then curly r which is distance from edge to P(new) becomes √ z^2 + (a/2)^2 as curly r is the hypotenuse. But when I use this relationship to try and forge something I actually get a wrong answer.. So I'm still not able to grasp the distance concept here😣)

[@zahidamona5798]

Good explanation sir 👍

[@zigzag_mode_on7140]

Well explained!... Thanku sir🌷

[@AnshulSharma1997]

Why have you takem the curly r, exactly at the distance a/2 from square and not anywhere else. For me, it is like for a point in a wire only you are finding the electric field.

[@brandonberisford]

I'm not sure I understand your question, I'm sorry could you elaborate a bit and I will try to answer the best I can.

[@mnakooda489]

in the textbook it says x runs from 0 to L , so shouldn't it be integrated from 0 to L

[@brandonberisford]

I moved my origin to the center of of one of the sides of the square, which is why I've integrated from -L to L.

[@Hkazmi25]

it should be Z dash.. when you plugin the values..

[@luishenriquebiassio7791]

Bons videos

[@gosnfrenki]

Hello sir, please can you answer me why we didnt take dE=2 times dEcos theta? Why just once?

[@brandonberisford]

Because we're just looking at the electric field contribution from a single differential piece of the line charge.

[@brandonberisford]

I showed two portions of it to show that the situation is symmetrical.

[@gosnfrenki]

Thank you sir!!