Question

Two identical charges , $q_1$ are placed at the vertices of a right triangle as shown above.
If a third charge $q_2$ is placed at the vertex of angle formed by the two legs, what is the magnitude of the net electrostatic force acting on $q_2$?

• A. $\frac{2k{q}_1{q}_2}{5}$
• B. $\frac{5k{q}_1{q}_2}{4}$
• C. $\frac{\sqrt{17}k{q}_1{q}_2}{4}$
• D. $\frac{17k{q}_1{q}_2}{4}$
• E. $\frac{5k{q}_1{q}_2^2}{4}$

Answer 1

Answer: (C)

$\frac{\sqrt{17}k{q}_1{q}_2}{4}$

According to Coulomb's law ,

force on $q_2$ by $q_1$ at a distance of 2cm is ,

$F_1=k{q}_1{q}_2/2^2$

or $F_1=k{q}_1{q}_2/4$

force on $q_2$ by $q_1$ at a distance of 1cm is ,

$F_2=k{q}_1{q}_2/1^2$

or $F_2=k{q}_1{q}_2/1$

the resultant of these forces at $q_2$ is

$R=\sqrt{(k{q}_1{q}_2/4{)}^2+(k{q}_1{q}_2/1{)}^2}$

or $R=k{q}_1{q}_2\sqrt{17/16}=\sqrt{1}7k{q}_1{q}_2/4$