In the figure three identical conducting spheres form an equilateral triangle of

Question

In the figure three identical conducting spheres form an equilateral triangle of side length d = 16.2 cm. The sphere radii are much smaller than d and the sphere charges are qA = -1.65 nC, qB = -4.69 nC, and qC = +9.42 nC. (a) What is the magnitude of the electrostatic force between spheres A and C? The following steps are taken: A and B are connected by a thin wire and then disconnected; B is grounded by the wire, and the wire is then removed; B and C are connected by the wire and then disconnected. What now are the magnitudes of the electrostatic force (b) between spheres A and C and (c) between spheres B and C?

Explanation / Answer

as the sphere's radius is very less compared to d, it can be assumed to be a point charge.

a) electrostatic force between any two charged particle q1 and q2 seprated by distance x is given by

9*10^9*q1*q2/x^2

the force is attractive in nature if charges are of opposite sign

force is repulsive in nature if charges are of same sign.

sphere A and sphere C have opposite charges.

hence the force will be attractive in nature i.e. force on A will be towards C and away from A and similarly, force on C will be towards A and away from C.

force magnitude=9*10^9*qA*qC/d^2=5.33*10^(-6) N

part b.

step 1: A and B are conencted by a thin wire.

the thin wire being a conductor it will bring the charge on both the spheres to same value.

hence charge on A=charge on B=(qA+qB)/2=-3.17 nC

step 2: B is grounded by the wire. then B has voltage=0 and it is discharged of all the charge.

hence charge on B=0 C

step 3: B and C are now connected.

then as observed in step 1, the charge on B=charge on C=(0+9.42)/2=4.71 nC

so after all three steps,

new charge values are:

qA=-3.17 nC
qB=4.71 nC
qC=4.71 nC

hence force between A and C=9*10^9*qA*qC/d^2=5.12*10^(-6) N,

and the force is attractive in nature.

c) force between B and C=9*10^9*qB*qC/d^2=7.6077*10^(-6) N

and the force is repulsive in nature.

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