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Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss's law. (Hint: Use Coulomb's law directly and evaluate the necessary integral.) |
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Answer» Solution :Consider INFINITELY extended line of charge with constant linear charge density `lambda` . Let us determine its electric field at perpendicular distance r say at point P as shown in the figure. For this consider a charge element dq at distance y from central point O. Here, `dq = lambdady` `(therefore lambda = (dq)/(dy))` MAGNITUDE of Coulombian force exerted by above charge element dq on test charge `q_(0)`placed at point P is, `dF = k((dq)q_(0))/(CP)^(2) = (k(dq)q_(0))/(y^(2) + r^(2))`........(1) Magnitude of electric field of charge element dq at point P is, `dE = (dF)/(q_(0)) = (kdq)/(y^(2) + r^(2))`.........(2) Similary magnitude of electric field of charge element dq (selected on OPPOSITE SIDE of central point O) at point P is also, `dE = (kdq)/(y^(2) + r^(2))` When we resolve above two electric fields into two mutually perpendicular components at point p we find that dEsinG components cancel each other (because of equal magnitudes and opposite directions). Hence, we get electric field at point P only due to `dEcostheta`components which are in same directions and so we can make their scalar addition to find resultant electric field E at point P Thus, `E = int dEcostheta` `int(kdq)/(y^(2) + r^(2)).r/sqrt(y^(2) + r^(2))` From eqn. (2) and from `DeltaCOP` `therefore E = intklambda dy xx r/(y^(2) + r^(2))^(3//2)`...........(3) In right angled `triangleCOP` `tantheta = y/rrArr y = r tan theta`.......(4) For a given point P value of r is constant and so from above equation, `(dy)/(d theta) = r(d)/(d theta) (tan theta)` `therefore (dy)/(d theta) = r sec^(2)theta` `therefore dy = rsec^(2)theta d theta`........(5) Now, `(y^(2) + r^(2))^(3//2) = (r^(2)tan^(2)theta + r^(2))^(3//2)` (From equation (4)) `=(r^(2) sec^(2)theta)^(3//2)` `=r^(2)sec^(3)theta`.........(6) From equation (4) `tan theta = y/r` When `y =-infty, tan theta = -infty rArr theta = -pi/2` `y=+infty, tan theta =+ infty rArr theta = + pi/2` ............(7) From eqn (3), (5),(6) and (7), `E = int_(-pi//2)^(+pi//2) klambda(r sec^(2)theta d theta) xx r/(r^(3) sec^(3)theta)` `therefore E=(klambda)/r {sin(pi/2)- sin(-pi/2)}` `therefore E =(2klambda)/r` `therefore E = (lambda)/(2pi epsilon_(0)r) (therefore k=1/(4pi epsilon_(0)))` |
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