Question

5. Two particles carrying a charge of qi +8q and qp -2q are fixed in place: charge qi is at the origin while the charge q2 is at a distance R. A what point (other than infinitely away) can a proton be placed so that it is in equilibrium (the net force on it is zero)? Is that equilibrium stable or unstable? (That is, if the proton is displaced, do the forces drive it back to the point of equilibrium or drive it further away? 92 q1

Answer 1

Let the charge of the proton is +q and it will be in position x from the origin where the net force on it is zero. Thus the force due to q1 and q2 should be equal in magnitude and opposite in direction. As the charges q1 and q2 are of opposite signs the proton cannot be between the two charges. Since q1 has larger magnitude of charge thus the charge should be placed closer to q2.

Let the charge of the proton is + q and it will be in position x from the origin where the net force on it is zero. Thus the force due to q1 and q2 should be equal in magnitude and opposite in direction. As the charges q1 and q2 are of opposite signs the proton cannot be between the two charges. Since q1 has larger magnitude of charge thus the charge should be placed closer to q2. The force due to q_1 F_1 Vector = k q_1 q/x^2 = 8k q_2/x^2 i^hat [Therefore q_1 = 8q] The force due to q_2 F_2 Vector = k q_2 q/(x - R)^2 = -2k q^2/(x - R)^2 i^hat [Because q_2 = -2q] Therefore F_1 Vector + F_2 Vector = 0 = > 8k q^2/x^2 - 2k q^2/(x - R)^2 = 0 \r\n = > 8/x^2 = 2/(x - R)^2 = > 4/x^2 = 1/(x - R)^2 Taking squareroot on both sides we get

Since the three charges are in equilibrium due to electrostatic force only therefore the equilibrium is unstable by Earnshaw's theorem which states that a group of charges interacting solely due to electrostatic force cannot be in stable equilibrium.