Summary - Physics

What is the magnitude of the force between two point charges of (+3 \\\\mu C) and \\\\(-5 \\\\mu C), separated by a distance of 2 meters in vacuum? Use Coulomb\\\'s Law for your calculation.

Answer: Coulomb\\\'s Law states that the force (F) between two charges (q_1) and (q_2), separated by a distance (r), is given by: [ F = k \\\\frac{|q_1 q_2|}{r^2} ] Where: - (k) is Coulomb\\\'s constant ((8.988 \\\\times 10^9 , \\\\text{N m}^2/\\\\text{C}^2)) - (q_1 = +3 \\\\mu C = 3 \\\\times 10^{-6} , C) - (q_2 = -5 \\\\mu C = 5 \\\\times 10^{-6} , C) - (r = 2 , m) Plugging in the values: [ F = 8.988 \\\\times 10^9 , \\\\text{N m}^2/\\\\text{C}^2 \\\\times \\\\frac{(3 \\\\times 10^{-6} , C)(5 \\\\times 10^{-6} , C)}{(2 \\\\, m)^2} ] [ F = 8.988 \\\\times 10^9 \\\\times \\\\frac{15 \\\\times 10^{-12} , C^2}{4 , m^2} ] [ F = 8.988 \\\\times 10^9 \\\\times 3.75 \\\\times 10^{-12} , \\\\text{N} ] [ F = 33.705 , \\\\text{N} ] Thus, the magnitude of the force between the two charges is approximately (33.7 , N).

A small object carrying a charge of (2 \\\\mu C) is placed at a point (A) in vacuum, and experiences a force of (0.3 N) directed towards another point (B). Calculate the magnitude of the electric field at point (A) due to the charge causing the force.

Answer: To find the electric field (E) at point (A) due to the other charge at point (B), use the relation between the force (F) experienced by a charge (q) in an electric field (E): [ E = frac{F}{q} ] Given, ( F = 0.3 , N ) and ( q = 2 \\\\mu C = 2 \\\\times 10^{-6} C ), [ E = frac{0.3 , N}{2 \\\\times 10^{-6}, C} = 150,000 , \\\\text{N/C} ] Thus, the magnitude of the electric field at point (A) is (150,000 , N/C).

A point charge of (+4 mu C) is at the origin, and a second point charge of (-2 mu C) is located at (x = 3 , m). What is the electric potential energy of the system?

Answer: The electric potential energy (U) of a system of two point charges is given by the formula: [ U = k frac{q_1 q_2}{r} ] Where: - (k) is the electrostatic constant, (8.988 \\\\times 10^9 , {N m}^2 {C}^2) - (q_1 = +4 mu C = 4 \\\\times 10^{-6} , C) - (q_2 = -2 mu C = -2 \\\\times 10^{-6} , C) - (r = 3 , m) Plugging in the values: [ U = 8.988 \\\\times 10^9 , {N m}^2 {C}^2 \\\\times frac{(4 \\\\times 10^{-6} , C)(-2 \\\\times 10^{-6} , C)}{3 , m} ] [ U = 8.988 \\\\times 10^9 \\\\times frac{-8 \\\\times 10^{-12} , C^2}{3 , m} ] [ U = -23.9632 \\\\times 10^6 , {J} ] [ U approx -24.0 , {J} ] Thus, the electric potential energy of the system is approximately (-24.0 , J).

What is Coulomb\\\'s Law and what does it describe?

Answer: Coulomb\\\'s Law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is given by ( F = k frac{|q_1 q_2|}{r^2} ), where (F) is the magnitude of the force, (q_1) and (q_2) are the charges, (r) is the distance between the charges, and (k) is the Coulomb constant. This law describes the force as being attractive if the charges are of opposite signs and repulsive if the charges are of the same sign.

What is the principle of superposition in electrostatics?

Answer: The principle of superposition in electrostatics states that the total electric force acting on a specific charge due to several other charges is the vector sum of the individual forces that each of the other charges alone would exert on the specific charge. This principle is critical because it allows for the analysis of the electric forces in a system containing multiple charges by considering the contribution from each charge independently. The resultant force vector is simply the sum of these individual vectors, regardless of the presence of other charges. This principle holds true for both electric fields and electric potentials, where the total electric field or potential at a point is the sum of the fields or potentials due to each charge considered separately.