Problems on Equilibrium and Lami's Theorem

PROBLEMS ON EQUILIBRIUM AND LAMI'S THEOREM Equilibrium is a state for a mass/charge/particle when its state of motion remains unaffected, which implies that the net force acting on it is zero. Since the force is interconnected with the field, the previous statement also implies that the net field acting has to be zero. Types of equilibrium Stable equilibrium: In this case, if a particle is subjected to an impactive force, it tries to go back to its initial state of motion. Unstable equilibrium: In this case, if a particle is subjected to an impactive force, it cannot go back to its initial state of motion. Neutral equilibrium: In this case, if a particle is subjected to an impactive force, it gets displaced, but its current and initial states of motion have no difference. Stable And Unstable Equilibrium [Image of two positive charges (+q) on either side of a negative charge (-q). The system is labeled STABLE.] When a charged particle experiences a slight displacement from its equilibrium position, and the resultant force restores it back to the original position, it is considered to be in stable electrostatic equilibrium. In this scenario, if we slightly displace the -q charge perpendicular to the line connecting the two +q charges, it will tend to return to its initial position. The combined electrostatic force generated by the two positive charges will pull it back towards the initial position. [Image of three positive charges (+q) in a line. The system is labeled UNSTABLE.] If a charge particle undergoes a slight displacement from its equilibrium position and the resting force attempts to push the particle further from the equilibrium position, it is considered to be in an unstable electrostatic equilibrium. In this instance, since all three charges are negative, the collective force repels the charge -Q away from its initial position. [Image of two negative charges (-q) on either side of a positive charge (+q). The system is labeled STABLE.] In this scenario, if we make a slight displacement of the +Q charge perpendicular to the line connecting the two -q charges, it will be pulled back to its original position. The combined electrostatic force generated by the two negative charges will pull it back towards the initial position. Since the charge being considered has the opposite sign compared to the other two charges, the resulting net force will be attractive. [Image of two positive charges (+q) on either side of a negative charge (-q). The system is labeled UNSTABLE.] In this scenario, if we slightly displace the +Q charge along the line connecting the two -q charges, it will move in the direction of the displacement. As the distance between them diminishes, the repulsive force will increase, causing it to accelerate away from the equilibrium point. The two negative charges will act in the direction of the displacement. Thus, the charge will be in an unstable equilibrium. If all three charges have the same sign (i.e., all negative or all positive), and we slightly displace the central charge in either direction along the line connecting the other two charges, the center charge will undergo simple harmonic motion (SHM). Pendulum Problems At equilibrium Balancing the forces in y-direction: = Tcosθ = mg Balancing the forces in x-direction: = Tsinθ = Fₑ = (kq₁q₂)/r² Ex. Find the mathematical relation between θ₁ and θ₂ given that q₁ > q₂? (a) θ₁ > θ₂ (b) θ₁ > θ₂ (c) θ₁ = θ₂ (d) θ₁ ≥ θ₂ Sol. As you can see, the electrostatic force acting on these two charges is same. The length of the string and their masses are equal. So the angle by which they are rotated from the mean position will be equal. The correct answer is (c) θ₁ = θ₂ Ex. Find the mathematical relation between θ₁ and θ₂ given that m₁ > m₂ (a) θ₁ > θ₂ (b) θ₁ = θ₂ (c) θ₁ < θ₂ (d) θ₁ ≥ θ₂ Sol. As you can see, the electrostatic force acting on these two charges is same. The length of the string is equal for both charges. But their mass is different. So the heavier charge will shift more towards the mean position. Hence it will have smaller angle subtended with the normal to the mean position. The correct answer is (c) θ₁ < θ₂ Pendulum Problems At equilibrium Fₑ = w tanθ Tcosθ = mg Tsinθ = Fₑ = (k(d-x)²)/x² PBD of a charge at equilibrium, LAMI'S Theorem When three coplanar, concurrent forces act on a body to maintain it in static equilibrium, each force is directly proportional to the sine of the angle formed by the other two forces. F₁/sin α = F₂/sin β = F₃/sin γ Ex. A charge +q of mass m is attached to a massless string of length l. the string is hinged at point O and is free to rotate about the hinge. Another charge +q is placed on the vertical wall at a depth of l from point O. Find the tension (T) in the string when the system reaches its equilibrium state. Sol. From the triangle: θ + 2x = π As the system is at equilibrium. So, by using Lami’s theorem: Fₑ/sin(π-x) = mg/sin(π-x) = T/sin(θ) Fₑ/sin x = mg/sin x = T/sin(π-2x) = T/sin(2x) = T = mg Layman's View of Geography of Electric Charges: The electric charges also possess different states of equilibrium. To begin the discussion, we should first examine the case of a positive charge. Let's begin with a positive charge. We have a positive charge. When we talk about positive and negative charges, we can consider them on the same level, as shown in the figure: However, if the concept of field lines is incorporated with positive and negative charges, then the position of the positive charge can be thought of as the top of a hill from which a fountain originates. It is because field lines originate from the positive charge. The position of the negative charge can be thought of as a valley or the bottom of a tunnel, because field lines originate from the positive charge and end at the negative charge. Hence, it can be said that the positive charge acts as the source and the negative charge acts as the sink. Therefore, it can be said that from the side view/topical view, the positive charge and the negative charge are not at the same level. The side view of the geography of the charges are shown below. Equilibrium of electric charges: When two positive charges of an equal magnitude are placed side by side, from layman's view of the geography of charges, they can be seen as shown in the figure. From the figure, the position of two positive charges of equal magnitude can be thought of as two different peaks of a hill or the same height. Now, the question arises, what is the possibility of equilibrium in this arrangement? Yes, the valley or region between the two peaks is the stable equilibrium position. Suppose that two positive charges of unequal magnitude are placed side by side. For this case as well, the valley or region between the two peaks is at the stable equilibrium position; but the region becomes smaller here and is closer to that charge with less magnitude. When two negative charges of equal magnitude are placed side by side, from layman's view of the geography of charges, the position of the two negative charges of equal magnitude can be thought of as a valley of flatness of the same depth. Now the question arises, what is the only possibility of equilibrium in this arrangement? Yes, the region (the peak) between the two charges is the unstable equilibrium position. Null Point: The null point is a position where the net field turns out to be zero. E_net is a vector quantity. Consider two point charges q1 and q2 as shown in the figure. For the net field to be zero at point P, the following should be true: E_net = 0 E₁ + E₂ = 0 [The negative sign arises because the directions of E₁ and E₂ are opposite. 1/(4πε₀) * (q₁/r₁²) - 1/(4πε₀) * (q₂/r₂²) = 0 ] q₁/r₁² = q₂/r₂² Ex. Find the position along the line joining two point charges where the net electric field is zero. Sol. In this case, the electric field will be zero at a point P, which is a distance away from the charge on the left as shown in the figure. Since both the charges are positive, the direction of the electric field on P due to these charges will be opposite. Also, we assumed the net electric field to be zero. E_net = 0 |E₁| = |E₂| 1/(4πε₀) * q/x² = 1/(4πε₀) * q/(d-x)² x² = (d-x)² x = ±(d-x) x = d/2 [By taking only the positive value] Therefore, the equilibrium position or the null point will be at the middle of the line joining the two equal positive charges. Ex. Find the position along the line joining two point charges where the net electric field is zero. Sol. Let the electric field be zero at point P, which is x distance away from the charge at the left. Since both the charges are positive, the direction of the electric field at P due to these charges will be opposite. Also, we assumed the net electric field at point P to be zero. E_net = 0 |E₁| = |E₂| 1/(4πε₀) * q/x² = 1/(4πε₀) * 4q/(d-x)² x²/(d-x)² = 1/4 x/(d-x) = ±1/2 2x = ±(d-x) [By taking only the positive value] 2x = d-x x = d/3 Therefore, the equilibrium position or the null point will be closer to the smaller positive charge along the line joining the two unequal positive charges.