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Find the electric field intensity at a point P which is at a distance R (point lying on the perpendicular drawn to the wire at one of its end) from a semi-infinite uniformly charged wire. (Linear charge density = lambda). |
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Answer» SOLUTION :Field at point P is due to an ELEMENTAL charge. `dq(=lambdadl)` `dE=k(dq)/((R^(2)+L^(2)))` The COMPONENT along x-axis `dE_(x)=(kdqcostheta)/((R^(2)+l^(2)))=(k(dq)R)/((R^(2)+l^(2))^(3//2))` `therefore E_(x)=underset(0)overset(oo)int(1)/(4piepsilon_(0))(lambdadlR)/((R^(2)+l^(2))^(3//2))impliesE_(x)=(lambda)/(4piepsilon_(0)R)` Similarly, `E_(y)=underset(0)overset(oo)int(1)/(4piepsilon_(0))(lambda(dl)l)/((R^(2)+l^(2))^(3//2))` `therefore E_(y)=(lambda)/(4piepsilon_(0)R)` `therefore E=sqrt(E_(X)^(2)+E_(Y)^(2))=(lambda)/(2.sqrt(2)piepsilon_(0)R),tantheta=(E_(y))/(E_(x))` `implies theta=45^(@)` 
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