InterviewSolution
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InterviewSolution
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Obtain the formula for the electric field due to a long thin wire of uniform linear charge density lambda without using Gauss's law. |
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Answer» Solution :Let us consider a long thin wire of linear charge density `lambda`. We have to find the RESULTANT electric field due to this wire at point P. Now, consider a very small element of length dx at a distance x from C.  The charge on this ELEMENTARY portion of length dx `q = lambda dx` ...(1) Electric field intensity at point P due to the elementary portion `DE=1/(4pi epsi_(0)). q/((OP)^(2))=1/(4pi epsi_(0)) (lambda dx)/((OP)^(2))` [` :'` from (1)] Now, in `Delta PCO""(PO)^(2)=(PC)^(2)+(CO)^(2)` `(OP)^(2)=r^(2)+x^(2)` `dE=1/(4pi epsi_(0)) (lambda dx)/((x^(2)+r^(2)))` ...(2) The components of dE are `dE cos theta` along PD and `dE sin theta` along PF. Here, there are so many elementary portion. So all the `dE sin theta` components balance each other. the resultant electric field at P is due to only `dE cos theta` components. The resultant electric field due to elementary component, `dE'=dE cos theta` `dE'=1/(4pi epsi_(0)). (lambda d x)/((x^(2)+r^(2))) cos theta` ...(3) In `Delta OCP tan theta=x/r implies x =r tan theta` Differentiation with respect to `theta`, we get `dx= r sec^(2) theta d theta` Putting in equation (3), we get `dE'=1/(4pi epsi_(0)) (lambda.r sec^(2) theta d theta cos theta)/((r^(2)+r^(2) tan^(2) theta))` `dE'=1/(4pi epsi_(0)) (lambda.r sec^(2) theta d theta cos theta)/(r^(2) sec^(2) theta)` `dE'=1/(4pi epsi_(0)). lambda/r cos theta d theta` As the wire os of infinite length, so integrate WITHIN the limits `-pi/2` to `pi/2`, we get `E'=int dE'=1/(4pi epsi_(0))lambda/r underset(-pi//2)overset(pi//2)(int)cos theta d theta` `E'=1/(4pi epsi_(0)).lambda/r[sin theta]_(-pi//2)^(pi//2)=1/(4pi epsi_(0)) lambda/r ["sin"pi/20sin(-pi/2)]` `E'=1/(4pi epsi_(0)) lambda/r [1+1]` `E'=(2 lambda)/(4pi epsi_(0) r)` `:. E'=lambda/(2pi epsi_(0) r)` |
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