For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

#tag_title#Step 2: Calculate the force acting on q due to one charge Q#tag_content# Next, let's calculate the force acting on q due to one of the charges Q. We can use Coulomb's law to find this force: \(F = k\frac{Qq}{r^2}\) where F is the force, k is the electrostatic constant, Q is the charge at one vertex, q is the charge at the center, and r is the distance between the two charges. In our case, the distance r is equal to the diagonal of half of the square. Using the Pythagorean theorem, we get: \(r = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2}\) \(r = \sqrt{a^2}\) \(r = a\) Now we can substitute this distance into the force equation: \(F = k\frac{Qq}{a^2}\) #tag_title#Step 3: Calculate the net force acting on q#tag_content# Now, we need to calculate the net force acting on q due to all the charges Q. Since the charges Q are symmetrically placed, the forces acting on q will be at 45 degrees to the x and y axes. We can decompose these forces into their x and y components and sum them up: \(F_{x, net} = F\cos(45^\circ) + F\cos(45^\circ)\) \(F_{y, net} = F\sin(45^\circ) + F\sin(45^\circ)\) Substitute the force value calculated in step 2: \(F_{x, net} = 2\left(k\frac{Qq}{a^2}\right)\cos(45^\circ)\) \(F_{y, net} = 2\left(k\frac{Qq}{a^2}\right)\sin(45^\circ)\) Since the net force on q is zero, both the x and y components of force must be equal to zero: \(F_{x, net} = 0\) \(F_{y, net} = 0\) #tag_title#Step 4: Solve for q#tag_content# Finally, we can solve for q using the equations from step 3. We can use either the x or y component equation, because they both contain q and will give us the same result: \(0 = 2\left(k\frac{Qq}{a^2}\right)\cos(45^\circ)\) Now solve for q: \(q = -2\sqrt{2}Q\) The value that makes the resultant force on q zero is: \(q = -2\sqrt{2}Q\) Therefore, the answer is (A) -2√2Q.