JEE MAIN - Physics (2005 - No. 44)
Two point charges $$+8q$$ and $$-2q$$ are located at $$x=0$$ and $$x=L$$ respectively. The location of a point on the $$x$$ axis at which the net electric field due to these two point charges is zero is
$${L \over 4}$$
$$2$$ $$L$$
$$4$$ $$L$$
$$8$$ $$L$$
Explanation
$${{ - K2q} \over {{{\left( {x - L} \right)}^2}}} + {{K8q} \over {{x^2}}} = 0 \Rightarrow {1 \over {{{\left( {x - L} \right)}^2}}} = {4 \over {{x^2}}}$$
or, $${1 \over {x - L}} = {2 \over x} \Rightarrow x = 2x - 2L$$
or, $$x=2L$$
or, $${1 \over {x - L}} = {2 \over x} \Rightarrow x = 2x - 2L$$
or, $$x=2L$$
Comments (0)
