00:01
Hi, so well in this question we have given a figure and and we need to find the magnitude in the direction of the electric field.
00:09
So first of all we will draw the diagram okay.
00:11
So these are the diagram like this is x -axis over here, this is x -axis and this is y -axis, okay? and the curve is going like this.
00:24
This is just like the quarter circle and one more over here and one more over there, okay? and charge on this curve is plus q, this curve is minus 4 q, this curve is plus 9 q, okay? and the radius of this much is r, this one is 2r, and this one is 3r, okay? now we need to find, since over here we have given the q equals to, that is 2 q4 the q stand for, q stand for 2 microculum, okay? and one more thing, a radius r equal to 10 cm, which is also given as a 0 .1, i think, i guess, yeah, 0 .1 meter.
01:09
Well, and what we have, q given and everything is given.
01:14
So we have to find the magnitude in the direction of the negative direction of the net electric field oriented to the arcs, okay? so since very simple question, we already know that the x will be the electric field due you this x component of this one one and two and three and the electric field in y direction will be one two and three okay so since our next go to each of the electric field so notice that the arc look at the same different in the only size okay okay so pi by two and pi so the only so we start by integrating x okay so x will be nothing but e cost theta how how so see we know that the left to feel due to this arc will be nothing but in this direction okay and due to the second one will be in this direction because this is negative charges and due to this third one that will be also in this direction okay and obviously since this is at the similar position that is 90 degree so this angle i think because these are due to the similarity or the similarity if this is a radius r and this is pi by two making angle so this will be just just just the mean mean between that the angle by sector of this one will be the line this angle will be 45 degree obviously so x will be e cost theta d s okay so e x net e x will be nothing but e e first one x plus e second one minus of obviously this will be in the negative region e2x plus of e3x because this this have the negative charges and similarly for y e1y minus of e2 y plus of e3y okay so since we know that e x will be nothing but e of cost theta and d s okay okay electric field will be nothing but we already know that k upon lambda lambda is nothing but the charge per unit length okay upon r square distance from there center to the point and we need to find the electric field at what electric field at on the origin obviously so this is the distance are r that means the same from the set their center i mean their surface to the origin or the point where we need to find electric field okay into cost theta into ds so now we know that where lambda is our clients everything we have been sent over here so will be nothing but k lambda pi by 2 to pi because this angle is pi by 2 over here and this pi pi pi by 2 to pi and k lambda upon r square cost theta d theta d s obviously over there in terms of angle so this and these are constant and this will be r cost theta because we know that d s is equal to r d theta okay so differentiation of this one okay r d theta so that's why we have we can r d theta so r and r get cancelled out kl lambda by r is constant so we only with integration of post -teta we already know that is sine theta and from pi by two to pi that means k lambda upon r into this will be sine pi that means zero and minus sign of one minus of one okay so to complete the integration we expand lambda in its full form so lambda is equal to nothing but we already mentioned that charge per unit length that means u .m.
05:26
That means u .t .s.
05:31
Will be nothing but minus of lambda in place of lambda we calculate q and this is minus one obviously this is not lambda this is minus one so q upon theta r okay so, theta over here is nothing but pi by 2 because this is the one fourth of a circle and into r square one r is already over here earlier and similarly for ey, result will be same and why? because we have to put now e sine of theta ds, okay? so everything will be put it in k lambda upon r square sine theta and r d theta...