Consider four equal charges `q_1 , q_2 , q_3 and q_4 = q = + 1 muC` located at four different points on a circle of radius 1m, as shown in the figure. Calculate the total force acting on the charge q1 due to all the other charges.

Explanation

According to the superposition principle, the total electrostatic force on charge `q_1` is the vector sum of the forces due to the other charges,
`vec (F)_(1)^("tot") = vec( F)_(12) + vec( F )_(13) + vec( F)_(14)`
The following diagram shows the direction of each force on the charge `q_1`.

The charges `q_2 and q_4` are equip-distant from `q_1`. As a result the strengths (magnitude) of the forces `vec( F)_(12) and vec(F)_(14)` are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. the charge `q_3` is located farther compared to `q_2 and q_4`. Since the strength of the electrostatic force decreases as distance increases, the strength of the force `vec( F)_(13)` is lesser than that of forces `vec( F)_(12) and vec(F)_(14)`. Hence the vector representing the force `vec(F)_(13)` is drawn with smaller length compared to that for forces `vec(F)_(12) and vec(F)_(14)`.
`F_(13) = 2.25 xx 10^(-3)N`
`F_(12) = (kq^2)/( r_(21)^2) = F_(14) = (9 xx 10^9 xx 10^(-22))/(2)`
`4.5 xx 10^(-3)`N
From the figure, the angle `theta = 45^(@)`. In terms of the components, we have
`vec(F)_(12) = F_(12) cos theta hat (i) - F_(12) sin theta hat (j)`
`= 4.5 xx 10^(-3) xx (1)/( sqrt2) hat(i) - 4.5 xx 10^(-3) xx (1)/( sqrt2) hat(j)`
`vec(F)_(13) = F_(13) hat(i) = 2.25 xx 10^(-3) N hat(i)`
`vec(F)_(14) = F_(14) cos theta hat(i) + F_(14) sin theta hat(j)`
`=4.5 xx 10^(-3) xx (1)/(sqrt2) hat(i) + 4.5 xx 10^(-3) xx (1)/( sqrt2) hat(j)`
Then the total force on `q_1` is,
`vec(F)_(1)^("tot") = ( F_(12) cos theta hat(i) - F_(12) sin theta hat(j) ) + F_(13) hat(i)`
`+ (F_(14) cos theta hat(i) + F_(14) sin theta hat(j) )`
`vec (F)_1^("tot") = (F_(12) cos theta + F_(13) + F_(14) cos theta ) hat(i)`
`+ ( - F_(12) sin theta + F_(14) sin theta ) hat(j)`
Since `F_(12) = F_(14)`, the `j^(th)` component is zero. Hence we have
`vec(F)_(1)^("tot") = (F_(12) cos theta + F_(13) + F_(14) cos theta) hat(i)`
substituting the values in the above equation,
`= ((4.5)/( sqrt2) + 2.25 + (4.5)/( sqrt2) ) hat(i) = (4.5 sqrt(2) + 2.25 ) hat(i)`
`vec(F)_(1)^("tot") = 8.61 xx 10^(-3) N hat(i)`
The resultant force is along the positive x axis.