MCQ Solution: Electrostatics
Charges +9Q and -4Q are placed as shown in figure. The point at which electric field intensity is zero at a distance from B on the line joining AB will be :
1) l
2) 3l
3) 9 / 4l
4) 2l
Solution
Let x is distance from B where electric field is zero.
Electric field due to +9Q charge is E1 =k(9Q / x2)
Electric field due to −4Q charge is E2 = 
These fields are in same direction.
Net field is zero if E1+E2=0⇒
or,(3l−3x)2−(2x)2=0⇒(3l−3x+2x)(3l−3x−2x)=0
∴x=3l,
x=3l /5
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