Two charges 9e and 3e are placed at a distance r. what will be the distance of the point where the electric field intensity will be zero
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place a unit positive charge in between the charges
9 e and 3 e such that it is at a distance of x from the 3 e charge and (r-x) distance from the 9 e charge. As electric intensity is zero at the position of unit positive charge so K(1)(9e)/(r-x)^2= K(1)(3e)/(x^2). 3x^2=(r-x)^2. so (r-x)=√3x. r/(1+√3) =x.
9 e and 3 e such that it is at a distance of x from the 3 e charge and (r-x) distance from the 9 e charge. As electric intensity is zero at the position of unit positive charge so K(1)(9e)/(r-x)^2= K(1)(3e)/(x^2). 3x^2=(r-x)^2. so (r-x)=√3x. r/(1+√3) =x.
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