A 5.0-μC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-μC point charge is ...

electric field due to positive charge is away from it and electric field due to negative charge is towards it.


hence for x<0 ( x is the coordinate on x axis, assuming the meter stick is along x axis and zero point coincide with origin)

, the electric field due to +ve charge will be along -ve x axis and electric field due to

negative charge will be towards +ve x axis.

hence both the field will oppose each other.


for 0<x<0.5 m (ie. between the two charges ), electric field will be along the same direction i.e. +ve x axis.


again, for x>0.5 m, both the fields will oppose each other.


hence the zero electric field will occur either for x<0 or x>0.5 m

now, as we know for a charge Q, electric field at a distance d is given by

9*10^9*Q/d^2

hence field magnitude is directly proportional to charge and inversely proportional to square of the distance .

as 5 uC> 4 uC

, the distance for 5 uC has to be greater than distance for 4 uC so that both the fields will have same magnitude but opposite direction.


hence x>0.5 m

now,let zero field occurs at x=d


then 9*10^9*5*10^(-6)/(d+0.5)^2=9*10^9*4*10(-6)/(d-0.5)^2

==> 5*(d-0.5)^2=4*(d+0.5)^2

==> 5*(d^2-d+0.25)=4*(d^2+d+0.25)

==> d^2-9*d+0.25=0
solving for d , we get

d=8.972 m

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