Answer to Question #170894 in Physics for Arpit

Question

Two point charges 40uc and 10uc are separated by a distance of 0.9m in air. Find the position of point where resistant intensity is zero.

Answer 1

Let the distance between point charges $q_1$and $q_2$ be $r$. Let the position where the resultant electric field intensity is zero located at a distance $d$ from the $q_1=40\ \mu C$ charge and distance $(r-d)$ from the $q_2=10\ \mu C$ charge. Then, we can find this position by equating electric fields due to point charges $q_1$and $q_2$ and solving for $d$ :

$E_1=E_2,$ $\dfrac{kq_1}{d^2}=\dfrac{kq_2}{(r-d)^2},$ $\dfrac{d^2}{(r-d)^2}=\dfrac{q_1}{q_2},$ $\dfrac{d}{r-d}=\sqrt{\dfrac{q_1}{q_2}},$ $d=\dfrac{r}{3}\cdot\sqrt{\dfrac{q_1}{q_2}},$ $d=\dfrac{0.9\ m}{3}\cdot\sqrt{\dfrac{40\cdot10^{-6}\ C}{10\cdot10^{-6}\ C}}=0.6\ m.$

Therefore, the resultant electric field intensity is zero at 0.6 m away from the $40\ \mu C$ point charge and 0.3 m away from the $10\ \mu C$ point charge.