Question

Four charges each equal to q are placed at the four corners of a square and a charge q is placed at the center of the square.If the system is in equilibrium then the value of q is

Answer 1

Equilibrium is a state in which opposing forces are balanced. Since it is given that the system is in equilibrium that implies that all the forces on every charge are balanced and equal.

Let the given five charges be q1, q2, q3, q4 and q5

Given q1 = q2 = q3 = q4 = Q and q5 = q

Let the charges be placed along the square as shown in the figure.

Since four charges are placed at four corners of a square,

From below figure,

Let AB = BC = CD = DA = a

∴The diagonal DB of the given square = √2a  

and half of the diagonal, OB = a/√2

Considering the charge q2,

Let the force exerted on q2 by charge q1 be F1

Similarly, force on q2 by q3 be F3

force on q2 by q4 be F4

force on q2 by q5 be F5

We know the Coulomb’s law is F = qq0 / 4πεr^2  [where r is the distance between two charges]

∴ By symmetry, F1 = F3 = q1q2/4πεr^2 = q3q2/4πεr^2 = Q^2/4πεa^2

F4 = q2q4/4πεr^2 = Q^2/4πε(√2a)^2 = Q^2/4πε(2a^2)

F5 = q2q5/4πεr^2 = Q^2/4πε(a/√2)^2 = -2qQ/4πεa^2

Since the direction of forces on q2 are along the edges, as shown, their vector sum will lie along the diagonal from D to B and have a magnitude, (F1 cos45) + (F3 cos45)

Also, since the system is in equilibrium, the net force acting on q2 is zero

∴ (F1 cos45) + (F3 cos45) + F4 + F5 = 0

(Q^2/4πεa^2 * 1/√2) + (Q^2/4πεa^2 * 1/√2) + (Q^2/4πε(2a^2)) + (-2qQ/4πεa^2) = 0  [∵ cos45 = 1/√2]

On solving, we get

q = Q/4 * (1+2√2)

Answer 2

Answer:

If the system is in equilibrium then the value of q is $\mathrm{q}=-\frac{\mathrm{Q}}{4}(1+2 \sqrt{2})$

Explanation:

Step 1 : Evaluate the distances between the charges.

Consider the length of one side of the square be a.

Refer Diagram,

$In \triangle \mathrm{ACD},\\\\\mathrm{AD}^{2}+\mathrm{CD}^{2}=\mathrm{AC}^{2} \\\\\mathrm{a}^{2}+\mathrm{a}^{2}=\mathrm{AC}^{2} \\\\\mathrm{AC}=\sqrt{2} \mathrm{a}\\\\\mathrm{OC}=\frac{\mathrm{AC}}{2}=\frac{\mathrm{a}}{\sqrt{2}}$

So one side of the square be a/√2

Step 2 : Using the concept of Equilibrium of charge at center.

By symmetry, Force on charge at centre will be equal and opposite due to the diagonally opposite charges, which will cancel each other. Hence, Net force on central charge will always be zero, irrespective of value of charge q. So in order to find value of q, we have to check equilibrium of any one charge at corner.

Step 3 : Calculate Force due to all the charges at point C.

Force on charge at C due to B, $\quad F_{1}=\frac{K Q^{2}}{a^{2}}$

Force on charge at C due to D, $\quad F_{2}=\frac{K Q^{2}}{a^{2}}$

Force on charge at C due to A, $\quad F_{4}=\frac{K Q^{2}}{AC^{2}}$=$\frac{K Q^{2}}{2a^{2}}$

Force on charge at $\mathrm{C} due to \mathrm{q} at centre, \quad \mathrm{F}_{3}=\frac{\mathrm{KqQ}}{\mathrm{OC}^{2}}=\frac{2 \mathrm{KqQ}}{\mathrm{a}^{2}}$

Step 4 :Equilibrium condition at C :

For the system to be in equilibrium, net force acting on charge at C must be zero. So,

$\overrightarrow{\mathrm{F}}_{3}+\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{4}=0$

$Resultant of F_{1} and F_{2} (Along OC, by symmetry) =\sqrt{F_{1}^{2}+F_{2}^{2}}=\sqrt{2} F_{1} \\Since \left(\left|F_{1}\right|=\left|F_{2}\right|\right)$

Also,$F_{3} \& F_{4}$ are along OC, Therefore, magnitudes of sum of these forces should be zero.

$\quad\left|\mathrm{F}_{3}\right|+\sqrt{2}\left|\mathrm{~F}_{1}\right|+\left|\mathrm{F}_{4}\right|=0\\\Rightarrow \frac{2 \mathrm{KqQ}}{\mathrm{a}^{2}}+\sqrt{2} \frac{\mathrm{KQ}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{KQ}^{2}}{2 \mathrm{a}^{2}}=0\\\Rightarrow 2 \mathrm{q}=-\left(\sqrt{2} \mathrm{Q}+\frac{\mathrm{Q}}{2}\right)\\\Rightarrow \mathrm{q}=-\frac{\mathrm{Q}}{4}(1+2 \sqrt{2})$

If the system is in equilibrium then the value of q is $\mathrm{q}=-\frac{\mathrm{Q}}{4}(1+2 \sqrt{2})$