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Step-by-Step Solution
1. Understand the Problem
Four charges each of value −Q are placed at the corners of a square. A charge q is placed at the center of the square such that the entire system is in equilibrium. We need to find the value of q.
2. Label the Square and Charges
Let the square have side length a. Denote the four corners by A, B, C, and D, each carrying charge −Q. Let the center of the square be O, carrying charge q.
3. Distance from Center to Corner
The distance from the center O to any corner (say A) is half the diagonal of the square. Since the diagonal of a square of side a is
$ a\sqrt{2} $,
the distance OA is
$ \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}. $
4. Consider the Force on One Corner Charge
By symmetry, analyzing the net force on any one corner charge (e.g., the corner at A) suffices. For equilibrium, the net force on that corner charge must be zero. The forces on charge at A come from:
The other three −Q charges at B, C, D.
The charge q at the center O.
5. Net Force from the Other Corner Charges
Focus on corner A at (0,0). Place B, C, D accordingly (for instance, B at (a,0), C at (a,a), and D at (0,a)). The repulsive force contributions in the x-direction from charges at B and C (and similarly in the y-direction from D and C) can be summed up. Symbolically, the net x-component of force from the other three corners on A will be:
$ F_\text{x, (from corners)}
= - \frac{k Q^2}{a^2}
\;-\; \frac{k Q^2}{2\sqrt{2}\,a^2}. $
The negative sign indicates these forces act toward the negative x-direction (because all corner charges are −Q and repel each other).
6. Force on the Corner Charge from the Center Charge
Now consider the force due to charge q at O. Since A is negative (−Q) and q will turn out to be positive, they attract each other. Thus, the force on the corner charge is directed from A toward O (in the positive x and positive y directions if A is at (0,0)). The magnitude of this force is:
$ F_\text{AO}
= k \frac{|\!-Q \cdot q|}{\bigl(\frac{a}{\sqrt{2}}\bigr)^2}
= k \frac{Q\,|q|}{\frac{a^2}{2}}
= 2k \frac{Q\,|q|}{a^2}.
$
The x-component of this force is
$ \left(2k \frac{Q|q|}{a^2}\right) \times \frac{1}{\sqrt{2}}
= \sqrt{2}\,k \frac{Q|q|}{a^2}. $
Because the force is attractive and directed toward the center, it is positive in both x and y directions.
7. Impose Equilibrium in the x-Direction
For equilibrium, the sum of the x-components of all forces on the charge at A must be zero:
$ \Bigl(- \frac{kQ^2}{a^2} - \frac{kQ^2}{2\sqrt{2}a^2}\Bigr)
\;+\; \sqrt{2}\,k \frac{Qq}{a^2} = 0.
$
Cancel the common factor
$ \frac{kQ}{a^2} $
on both sides:
$ -Q \Bigl(1 + \frac{1}{2\sqrt{2}}\Bigr) \;+\; \sqrt{2}\,q = 0.
$
Hence,
$ \sqrt{2}\,q = Q \Bigl(1 + \frac{1}{2\sqrt{2}}\Bigr). $
Solving for q:
$ q = \frac{Q}{\sqrt{2}} \Bigl(1 + \frac{1}{2\sqrt{2}}\Bigr).
$
Recognize that
$ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, $
and simplify:
$ q = Q \left(\frac{\sqrt{2}}{2} + \frac{1}{4}\right)
= \frac{Q}{4}\bigl(2\sqrt{2} + 1\bigr)
= \frac{Q}{4}\bigl(1 + 2\sqrt{2}\bigr).
$
8. Final Answer
The charge at the center is:
$ \displaystyle q = \frac{Q}{4}\bigl(1 + 2\sqrt{2}\bigr).
$
