If you were to integrate the load intensity equations three...

Name: if you were to integrate the load intensity equations three times you will get the equation for 84558
Uploaded: 2024-03-21T15:16:20-08:00
Duration: 4 min 54 s
Channel: Supratim Pal
Description: if you were to integrate the load intensity equations three times you will get the equation for 84558

Transcript

00:01 Hello and welcome to this video solution of new mudded here we are given three charge and based on this you would find out the electric field intensity everywhere along the x -axis due to these three charges, right? so to the left of it there is this charge q and to the right again a charge q and the center you have got minus 4 q right and this is at x equal to 0 and here you have caught at a distance of x equal to a and this is also x equal to minus a right this is point a this is b and this is c right great so what we can do is we can start okay take certain points right let's say point one is at a region one at the right of the chart c region two in between c and b three is between a and b and four is to the left of a right so for region one you take a point here the electric field is pretty simple to calculate so ec is to the right because it's a positive charge similarly we have caught ea to the right and to the left will be eb right or e1 will be equal to ea plus ec minus eb right and let's say we have got r r to be greater than a right so electric field due to a will be equal to k times of ea is a and cr charge q right here ea sorry sorry k q by a is to the left right r r is from this center i take from center it's r right r plus a whole square plus ec is k q times of r minus a whole square minus eb is k times of 4q over r square right so this is the expression of electric field field right now e2 and obviously it will be directed what we see is mostly to the left because of high value of 4q right okay so e2 will be equal to here at the this is let's say at a distance r from the center at 2 what happens is electric field is directed to ebe is to the left from c again to the left ec and a to the right ea right so e e b plus ec minus ea right now this is equal to eb is now two means x will be equal to r and r should be less than a now eb is k q b q over b is at the center right so this is r square plus ec is k q by r plus a whole square and sorry ec is to the right of it right sorry then it will be a minus r right whole square minus minus sorry i just made a simple error for b the charge is 4q right 4q and this is kq by a minus r whole square minus here kq by a plus r whole square it will be right due to a right so this is 2 3 will be again same 3 will be again the same e3 will be equal to e2 right and fourth fourth one will be e4 will be equal to e1, right? again same, but the directions will be on the other side, right? i hope this is clear to you and a very good rest of the day...

Question

if you were to integrate the load intensity equations three times, you will get the equation for:

Question

if you were to integrate the load intensity equations three times, you will get the equation for:

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