Question

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 :

• A. $l$
• B. $3l$
• C. $\frac{9}{4}l$
• D. $2l$

Answer 1

Answer: (B)

$3l$

Let $x$ is distance from B where electric field is zero.
Electric field due to $+9Q$ charge is $E_1=k\frac{9Q}{x^2}$
Electric field due to $-4Q$ charge is $E_2=k\frac{-4Q}{(l-x{)}^2}$
These fields are in same direction.
Net field is zero if $E_1+E_2=0\Rightarrow k\frac{9Q}{x^2}-k\frac{-4Q}{(l-x{)}^2}=0$
or,$\frac{9}{x^2}-\frac{4}{(l-x{)}^2}=0$
or,$9(l-x{)}^2-4x^2=0$
or,$(3l-3x{)}^2-(2x{)}^2=0\Rightarrow (3l-3x+2x)(3l-3x-2x)=0$
$\therefore x=3l,x=\frac{3l}{5}$
Ans: (B)