Recent content by Rahulrj

Yes by symmetry x components cancel at the point, so are you agreeing with me that the R should be ##y \hat j##?

I don't know if I am confusing myself, according to the solution of the post #9, keeping line on the x axis, the denominator has to be just ##y \hat j## right? not ##x\hat i+y \hat j## because R there was a constant in the y coordinate. so from flux rule, the denominator 's' has to be ##y \hat...

Now that's confusing, In the above posts, I solved it keeping the line charge on x-axis and the result I got depended on the y-axis which I took to be R. In the first post, using the flux rule I got ##E = \frac{\lambda}{2\pi \epsilon_0 s}## and I took that s to be the given point which is...

Maybe that's where I am getting confused, the 'r' in the denominator represents the distance between point and wire. so when a point is given where field is to be calculated say ##r_*##, I should just use that point as the 'r' in the denominator correct?

I took it be along x-axis.

Okay that makes sense. So to make sure, if the point were ##r = 2xi +3yj##, I could use (0,3R) or if I take ##dy##, I could use (3R,0) right? so in this case, the denominator becomes ##(x^2+9R^2)##? In essence, we have to keep one of the coordinate as a constant, am i right? I can write it as...

So if this is right, don't we have to include the information of the point? up till here ##r=xi+yj## is not included in the integration,if I am correct. Also I am not really sure about using limits here, could you please hint me on that? and I did not understand the choice of ##\pi/4## too.

Oh sorry that was a typo, I didn't intend it to be like that. I can right away say that the x components cancel and y components add up. So now I have the point ##r=ix+jy## at a distance R (perpendicular length) from the line charge at x axis. I can draw a hypotenuse to an element ##dx## on the...

I did not get what you meant by changing variable. ##dl=dx## I am just not sure what ##r^2## here is and since integrating limits involve infinity, I am not certain on how to approach this problem in that form. I follow griffith's electrodynamics and in the example 2 of chapter 2 (mentioned so...

Homework Statement Find the electric field of an infinitely long straight wire of charge ## \lambda## C/m at a point ##r= ix+jy## Homework Equations ##\int E.da = \frac {Q}{\epsilon_0}## ##E= \frac{\int dq}{4\pi\epsilon_0 r^2} ##r The Attempt at a Solution Drawing a cylindrical gaussian...

Thank you very much! helped me fill lot of gaps in what I have learned.

In case b) I am not sure how to deduce from velocities but I can using acceleration, so the expression will be ##a_{1p} = a_{2p}+a_{21}## since observer 1 sees particle with uniform velocity ##a_{1p} = 0## and ##a_{21}=0## since observer 2 is moving at constant velocity wrt to 1 and therefore...

That's simple, I just have to take the derivative of v for acceleration. In the case of b, I am not sure how the same equation can apply they are taken wrt to star but the case is wrt to observer 1 making no mention of a star. So shouldn't the expression be (V of particle wrt obs 1 = V of...

Yes I figured that in your post #15. I was stuck when you said there is only one velocity of particle relative to star because I misunderstood taking it be that there is only one expression. So now it will be, ##v_2+v_{2p} = v_1+v_{1p}## and for the case in ##c##, ##v_1=v_2## and therefore...

That exactly was my doubt here that if I can write two equations of relative velocity wrt each observer as opposed to what you said that there is only one velocity of particle relative to star. So then the equation will be ##v_{rel1} = v_1+v_{1p}##