IIT JEE , AIEEE Forum - Physics , Electricity

sulekha_hi

Four equal positive charges (Q) are arranged at the four corners of a square of side a. A unit positive charge of mass m is placed at a height h above the centre of the square. What should be the value of Q in order that this unit charge may be in equilibrium?

elessar_iitkgp

If you draw the force vectors on the unit charge, it can be seen that the components of the forces parallel to the plane of the square cancel each other. The net force is then away from the plane of the square. Lets prove this:

In these types of problems, its best to use the vector equation approach. You can find the derivation of Coulomb's law in vector form in NCERT Text.
Coulomb's law in vector form is
F₁₂ = (q₁q₂(r₁ - r₂))/4

₀|(r₁ - r₂)|³where F₁₂ is the force on q₁ due to q₂
First mark the charges on the square as 1,2,3,4 and the unit charge as 5.
Now, assume the point of intersection of the diagonals as the origin, the diagonal through charge 1 to be the X axis and the diagonal through charge 2 to be the Y axis. Let the Z axis be from the center of the square, towards the unit charge.
Net force on the unit charge,
F₅ = F₅₁+F₅₂+F₅₃+F₅₄
= (1.Q(hk - (a/

2) i ))/4

₀((a²/2) + h²)^3/2 +(1.Q((hk - (a/

2) j )))/4

₀((a²/2) + h²)^3/2+(1.Q((hk + (a/

2) i )))/4

₀((a²/2) + h²)^3/2+(1.Q((hk + (a/

2) i ))/4

₀((a²/2) + h²)^3/2
= [Q/(4

₀((a²/2) + h²)^3/2)] x [hk - (a/

2) i + hk + (a/

2) i]
F₅ =[4Qh/(4

₀((a²/2) + h²)^3/2)]k
As we had suspected, the net force is directed away from the plane of the square. This force isn't restoring in nature, so we can rule out any speculation about SHM. For equlibrium,
F₅ = 0
Qh = 0
So, either Q = 0 or h = 0
Now, lets interpret these results.
If Q = 0, the unit charge is at equilibrium at any distance from the center of the square.
If h = 0, the unit charge is at equilibrium no matter what the value of Q is. That is, if the charge is placed at the center of the square, it remains at equilibrium for all values of Q. This is expected from elementary considerations of the orientations of the force vectors in this case.

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