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Two charges of 1.0xx10^(-6) coulombare separated bya distance 10 cm . Where will the electric field be zero on the line joining the two charges ? |
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Answer» Solution :This is clear that such a point should be some where between the CHARGES because in that case only , electric field will be in opposite direction . Suppose that pointis at a distance x from the charge `q_(1)` , THEREFORE this distance from the charge `q_(2)` will be `(10-x)`. `thereforeE_(1)=(1)/(4piepsilon_(0))(q_(1))/(x^(2))` and `E_(2)=(1)/(4piepsilon_(0))(q_(2))/((r-x)^(2))` But ACCORDING to the problem `E_(1)=E_(2)` `therefore(1)/(4piepsilon_(0))(q_(1))/(x^(2))=(1)/(4piepsilon_(0))(q_(2))/((r-x)^(2))` or , `(1.0xx10^(-6))/(x^(2))=(2.0xx10^(-6))/((10-x)^(2))` or , `(1)/(x)=(sqrt(2))/((10-x))`(Taking positive sign) |
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