Three small spheres each carrying a charge q are placed on the circumference of a circle of radius R, forming an equilateral triangle. If we place another charge Q at the centre of the circle, the force on Q will be :

Question

Three small spheres each carrying a charge q are placed on the circumference of a circle of radius R, forming an equilateral triangle. If we place another charge Q at the centre of the circle, the force on Q will be : (A) Zero (B)  (qQ/4 π ε0R2) (C)  (2qQ/4 π ε0R2) (D)  (3qQ/4 π ε0R2)

Tardigrade Admin · Accepted Answer

The resultant of three equal forces inclined to each other at 120° is zero. Because: F2 + F2 + 2F2 cos 120° = F2. That is the resultant of any two of them is equal and opposite to the third force.

Q. Three small spheres each carrying a charge $q$ are placed on the circumference of a circle of radius $R$ , forming an equilateral triangle. If we place another charge $Q$ at the centre of the circle, the force on $Q$ will be :

Zero

$\frac{qQ}{4 π ε _{0} R ^{2}}$

$\frac{2 qQ}{4 π ε _{0} R ^{2}}$

$\frac{3 qQ}{4 π ε _{0} R ^{2}}$

The resultant of three equal forces inclined to each other at $12 0^{\circ}$ is zero. Because : $F^{2} + F^{2} + 2 F^{2} cos 12 0^{\circ} = F^{2}$ . That is the resultant of any two of them is equal and opposite to the third force.

Q. Three small spheres each carrying a charge q are placed on the circumference of a circle of radius R, forming an equilateral triangle. If we place another charge Q at the centre of the circle, the force on Q will be :

Zero

4πε0​R2qQ​

4πε0​R22qQ​

4πε0​R23qQ​

The resultant of three equal forces inclined to each other at 120∘ is zero. Because : F2+F2+2F2cos120∘=F2. That is the resultant of any two of them is equal and opposite to the third force.

Q. Three small spheres each carrying a charge $q$ are placed on the circumference of a circle of radius $R$ , forming an equilateral triangle. If we place another charge $Q$ at the centre of the circle, the force on $Q$ will be :

$\frac{qQ}{4 π ε _{0} R ^{2}}$

$\frac{2 qQ}{4 π ε _{0} R ^{2}}$

$\frac{3 qQ}{4 π ε _{0} R ^{2}}$

The resultant of three equal forces inclined to each other at $12 0^{\circ}$ is zero. Because : $F^{2} + F^{2} + 2 F^{2} cos 12 0^{\circ} = F^{2}$ . That is the resultant of any two of them is equal and opposite to the third force.