Three charges `q_1=1muC, q_2=2muC` and `q_3=3muC` are placed on the vertices of an equilateral triangle of side 1.0 m. Find the net electric force acting on charge `q_1`

Answer

A, B, C, D

Explanation

Method I. In the figure
`
`|F_(1)|=F_(1)=1/(4piepsilon_0).(q_1q_2)/r^2`
=magnitude of force between `q_1 and q_2`
`((9.0xx10^9)(1.0xx10^-6)(2.0xx10^-6))/((1.0)^2)`
`=1.8xx10^-2N`
`Similarly `|F_(2)|=F_(2)=1/(4piepsilon_0). (q_1q_3)/r^2`
`=magnitude of force between `q_1` and `q_3`
`=((9.0xx10^9)(1.0xx10^-6)(3.0xx10^-6))/((1.0)^2)`
`=2.7xx10^-2N`
Now, `|F_"net")|=sqrt(F_1^2+F_2^2+2F_1F_2cos120^@)`
`=(sqrt((1.8)^2+(2.7)^2=2(1.8)(2.7)(-1/2)))xx10^-2N`
`=2.38xx10^-2N`
and `tan alpha=(F_2sin120^@)/(F_1+F_2cos120^@)`
`=((2.7xx10^-2)(0.87))/((1.8xx10^-2)+(2.7xx10^-2)(-1/2))`
or `alpha=79.2^@`
Thus, the net force on charge `q_1` is `2.38xx10^-2` N at an angle `alpha=79.2^@` with a line joining `q_1` and `q_2` as shown in the figure.
Method 2. In this figure method let us assume a coordinate axes with `q_1` at origin as shown in figure. The coordinate of `q_1,q_2` and `q_3` in this coordinate system are `(0,0,0), (1m,0,0)` and `(0.5m, 0.87m,0)` respectively. Now
`
`F_1`=force of `q_1` due to charge `q_2`
`=1/(4piepsilon_0) .(q_1q_2)/(|r_1-r_2|\^3)(r_1-r_2)`
`=((9.0xx10^9)(1.0xx10^-6)(-2.0xx10^-6))/((1.0)^3)[(0.1)hati+(0-0)hatj+(0-0)hatk]`
`=(1.8xx10^-2hati)N`
and `F_2`=Force on `q_1` due to charge `q_3`
`=1/(4piepsilon_0).(q_1q_3)/(|r_1-r_3|^3)(r_1-r_3)`
`=((9.0xx10^9)(1.0xx10^-6)(3.0xx10^-6))/(1.0)^3[(0-0.5)hati+(0-0-0.87)hatj+(0-0)hatk]`
`=(-1.35hati-2.349hatj)xx10^-2N`
Therefore net force on `q_1` is `F=F_1+F_2`
`=(0.45hati-2.349hatj)xx10^-2N`