Two identical conducting spheres, each of radius R, carry charges +Q and -Q/2 respectively. When the distance between their centres is d(d 2R), the attractive force between them is F. The spheres are now brought into contact and then separated to the initial distance d. The force between the spheres is nowa)attractive and of magnitude F/4b)repulsive and of magnitude F/4c)attractive and of magnitude F/8d)repulsive and of magnitude F/8Correct answer is option 'D'. Can you explain this answer?

Question

Two identical conducting spheres, each of radius R, carry charges +Q and -Q/2 respectively. When the distance between their centres is d(d > 2R), the attractive force between them is F. The spheres are now brought into contact and then separated to the initial distance d. The force between the spheres is nowa)attractive and of magnitude F/4b)repulsive and of magnitude F/4c)attractive and of magnitude F/8d)repulsive and of magnitude F/8Correct answer is option 'D'. Can you explain this answer?

Answer

Let's first find the electric field due to each sphere individually at the center of the other sphere.

For the sphere with charge Q, the electric field at the center of the other sphere is given by:

E = kQ/R^2

where k is the Coulomb constant.

For the sphere with charge -Q/2, the electric field at the center of the other sphere is given by:

E' = k(-Q/2)/R^2 = -(kQ)/2R^2

The total electric field at the center of the other sphere is the vector sum of these two fields. Since the spheres are identical, the direction of the electric field due to each sphere is along the line connecting their centers, and they are in opposite directions.

If we draw a diagram, we can see that the angle between the two electric fields is 60 degrees.

Using vector addition, we get:

E_total = sqrt(E^2 + E'^2 + 2EE'cos60)

E_total = sqrt[(kQ/R^2)^2 + (kQ/2R^2)^2 - (kQ^2/2R^4)]

E_total = (kQ/R^2) sqrt[1 + 1/4 - 1/2]

E_total = (kQ/R^2) sqrt(3/4)

E_total = (kQ/R^2) sqrt(3)/2

Now, the force between the two spheres is given by:

F = Q(E_total)

F = Q(kQ/R^2)(sqrt(3)/2)

F = kQ^2(sqrt(3)/2R^2)

Similarly, we can find the force between the two spheres when they are far apart (d >> R). In this case, we can treat each sphere as a point charge located at its center. The distance between the two point charges is d - 2R. The force between two point charges is given by:

F = kQ1Q2/(d - 2R)^2

where Q1 and Q2 are the charges on the spheres.

Substituting Q1 = Q and Q2 = -Q/2, we get:

F = kQ(-Q/2)/ (d - 2R)^2

F = -kQ^2/2(d - 2R)^2

Comparing this with the previous expression for F, we can see that the force decreases as (d - 2R)^-2 when the spheres are far apart.

Therefore, the force between the spheres decreases as (d - 2R)^-2 when they are far apart, and as (sqrt(3)/2R^2)Q^2 when they are close together.

Answer

When they were in contact they shared charge equally. So finally charge on both of them was q/4but before contact F=(kq.q/2)d^2and after contact F,=k(q/4*q/4)d^2soF,=F/8 and force in repulsive cause same type of charges repulse.

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