Two equal point charges Q = √2 μC are placed at each of the two opposite corners of a square and equal point charges q at each of the two corners. What must be the value of q so that the resultant force on Q is zero?

To find the value of qq such that the resultant force on each QQ is zero, we solve step by step.
Given:
1. Two charges Q=2 μCQ = \sqrt{2} \, \mu C are placed at two opposite corners of a square.
2. Two charges qq are placed at the other two corners of the square.
3. Side length of the square = aa.
4. The force on QQ due to the other charges must sum to zero.
Step 1: Forces acting on QQ
The QQ charge at one corner experiences three forces:
1. Force FQ−QF_{Q-Q} due to the other QQ (diagonally opposite).
2. Two forces FQ−qF_{Q-q} due to the qq charges (at adjacent corners).
Step 2: Expression for forces
1. Force between charges is given by Coulomb's law:
F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}
where k=14πϵ0k = \frac{1}{4\pi \epsilon_0} (Coulomb's constant), q1q_1 and q2q_2 are the charges, and rr is the distance between them.
2. Distance between the charges:
o Adjacent charges QQ and qq: r=ar = a.
o Diagonal charges QQ and QQ: r=2ar = \sqrt{2}a (Pythagoras theorem).
Step 3: Force due to the diagonal QQ
The force between the two QQ charges (diagonally opposite):
FQ−Q=kQ2(2a)2=k(2 μC)22a2=k2 μC22a2=kμC2a2.F_{Q-Q} = k \frac{Q^2}{(\sqrt{2}a)^2} = k \frac{(\sqrt{2} \, \mu C)^2}{2a^2} = k \frac{2 \, \mu C^2}{2a^2} = k \frac{\mu C^2}{a^2}.
This force acts along the diagonal of the square.
Step 4: Force due to adjacent qq charges
The force between QQ and qq (for each adjacent qq):
FQ−q=kQqa2.F_{Q-q} = k \frac{Qq}{a^2}.
These forces act along the sides of the square.
Step 5: Net force conditions
To make the resultant force on QQ zero:
1. The diagonal force FQ−QF_{Q-Q} must be balanced by the resultant of the two adjacent FQ−qF_{Q-q} forces.
2. The two adjacent forces FQ−qF_{Q-q} form a 90∘90^\circ angle. The resultant of these two forces is:
Fresultant of Q-q=(FQ−q)2+(FQ−q)2=2 FQ−q.F_{\text{resultant of Q-q}} = \sqrt{(F_{Q-q})^2 + (F_{Q-q})^2} = \sqrt{2} \, F_{Q-q}.
3. For equilibrium:
FQ−Q=Fresultant of Q-q.F_{Q-Q} = F_{\text{resultant of Q-q}}.
Substitute the expressions:
kμC2a2=2 (kQqa2).k \frac{\mu C^2}{a^2} = \sqrt{2} \, \left(k \frac{Qq}{a^2}\right).
Cancel kk and 1a2\frac{1}{a^2} on both sides:
μC2=2 Qq.\mu C^2 = \sqrt{2} \, Qq.
Substitute Q=2 μCQ = \sqrt{2} \, \mu C:
μC2=2 (2 μC) q.\mu C^2 = \sqrt{2} \, (\sqrt{2} \, \mu C) \, q.
Simplify:
μC2=2 μC q.\mu C^2 = 2 \, \mu C \, q.
Cancel μC\mu C (assuming μC≠0\mu C \neq 0):
q=μC2.q = \frac{\mu C}{2}.
Final Answer:
The value of qq must be:
q=μC2=22 μC.q = \frac{\mu C}{2} = \frac{\sqrt{2}}{2} \, \mu C.