Question

There are 4 charges at the corners of a square. q1 (top left corner)= q. q2 (top right corner)= Q. q3 (bottom left corner) = -q. q4= -Q. Take q = 1.02 µC and Q = 2.23 µC.) The side length of the square is 10 cm.
a) what are the magnitude and direction of the resultant force on q₁? (Take q₁ as the origin of the coordinate system and measure the angle counterclockwise from the positive x-axis, which is directed towards the right.)
b) What is the resultant force on the center of mass of the four charges?

Answer 1

Data Given

q = 1.02 uC, Q = 2.23 uC, L = 10 cm = 0.10 m

Refer to the diagram

Fox 10 cm FB Fc FNet FD 14.14 cm 10 cm -q -Q 10 cm

$F_{B} = \frac{kQq}{BA^{2}}= \frac{9\times 10^{9}\times 1.02\times 10^{-6}\times 2.23\times 10^{-6}}{(0.10)^{2}}= 2.047 N$  towards - X axis

$F_{D} = \frac{kqq}{AD^{2}}= \frac{9\times 10^{9}\times 1.02\times 10^{-6}\times 1.02\times 10^{-6}}{(0.10)^{2}}= 0.936 N$  towards -Y axis

$F_{C} = \frac{kqQ}{AD^{2}}= \frac{9\times 10^{9}\times 1.02\times 10^{-6}\times 2.23\times 10^{-6}}{(0.1414)^{2}}= 1.024 N$  Along the diameter AC

Now Net Force along X axis will be

$F_{x}= F_{Cx}-F_{B} = F_{C}\sin 45^{0}-F_{B}$

$F_{x}= 1.024 N\sin 45^{0}-2.047 N = -1.323 N$

Now Net Force along Y Axis

$F_{y}= -F_{Cy}-F_{D} = -F_{C}\cos 45^{0}-F_{D}$

$F_{y}= -1.024 N\sin 45^{0}-0.936 N = -1.66 N$

$F_{x}= 1.024 N\sin 45^{0}-2.047 N = -1.323 N$

Net Force at A will be

$F_{net} = \sqrt{F_{x}^{2}+F_{y}^{2}}= \sqrt{(-1.323)^{2}+(-1.66)^{2}}= 2.123N$

Direction will be

$\theta = 180^{0}+\tan^{-1}\frac{F_{y}}{F_{x}}= 180^{0}+\tan^{-1}\left ( \frac{-1.66}{-1.323} \right )$

$\theta = 180^{0}+51.45^{0} = 231.45^{0}$  Counterclockwise to + X axis.

Part B) There will be no force at the centre of mass of the four charges as there is no charge is placed.