Physics Answer Note #10

Question 1 The image formed by a thin convex lens when an upright object is placed between the focal point and the converging lens is A real, erect, diminished. B virtual, erect, diminished. C real, inverted, magnified. D virtual, inverted, magnified. Answer The correct answer is B, virtual, erect, diminished. Explanation: When an object is placed between the focal point and the converging lens, the image formed will be virtual, erect, and diminished. To understand why, we need to first understand the properties of a convex lens. A convex lens is a converging lens, meaning it brings parallel rays of light towards a focal point. This focal point is located on the principal axis of the lens, which is a line that passes through the center of the lens. Now, let's look at what happens when an object is placed between the focal point and the lens. The parallel rays of light from the object will pass through the lens and diverge outwards. As these rays continue to diverge, they will eventually intersect at a point behind the lens. This point is known as the virtual image. In terms of orientation, the image formed by a convex lens is always upright, which means it is erect. This can be understood by the fact that the rays of light from the object pass through the lens without crossing each other, and so the image is not inverted. Furthermore, the image formed is also diminished, which means it appears smaller than the actual object. This can be explained by the fact that the rays of light from the object are spread over a larger area as they pass through the lens, causing the image to appear smaller. Question 2 Geometric interpretation of the Kuhn Tucker Conditions in Utility Maximisation Problem, according to matrix notations Answer The Kuhn Tucker Conditions (KTC) are a set of necessary conditions for optimality in constrained mathematical optimization problems. These conditions play a pivotal role in understanding the relationship between unconstrained and constrained optimization problems. In the context of utility maximization, the KTC provide a framework for analyzing the trade-offs between utility maximization and the budget constraints of an individual. In this response, we will first define the KTC and then provide a geometric interpretation of these conditions using matrix notations. The Kuhn Tucker Conditions can be summarized as a set of four conditions: the positive Lagrange multipliers, the complementary slackness, the feasibility, and the nonnegativity. These conditions are necessary for a point to be an optimal solution in a constrained optimization problem. In the utility maximization problem, the objective is to maximize the utility function subject to the budget constraint. The Lagrange multiplier is a parameter that is associated with the budget constraint and represents the marginal utility gained from relaxing that constraint. The complementary slackness condition states that at the optimal solution, the product of the Lagrange multiplier and the budget constraint must equal zero. This condition shows the trade-off between the marginal utility gained from relaxation and the value of the budget constraint. The feasibility condition states that at the optimal solution, all constraints must be satisfied with equality. This condition mirrors the budget constraint and shows that the optimal solution must be on the boundary of the feasible region. The non-negativity condition ensures that all variables in the optimization problem are positive or zero, and thus restricts the feasible region to only positive values. To further understand the geometric interpretation of the KTC, we can use matrix notations. Let us consider a budget constraint of the form: p1x1 + p2x2 + ...+ pnxn = M where x1, x2,..., xn are the decision variables, p1, p2,..., pn are the prices of the goods, and M is the budget. This constraint can be represented in a matrix form as: Px = M where P is a vector of prices and x is a vector of decision variables. The Lagrange multiplier for this constraint is represented by λ and the gradient of the utility function with respect to x is represented by ∇ U. The KTC can then be stated as: 1. ∇ U - λP = 0 2. λ(M - Px) = 0 3. Px = M 4. λ ≥ 0 and x ≥ 0 Using these notations, the first two conditions can be visually represented as the intersection of the budget constraint line and the utility function curve at the optimal solution, as shown in the figure below. [insert figure here] The third condition represents the optimal solution lying on the budget constraint, whereas the fourth condition restricts the feasible region to only positive values. Question 3 Given point charges that are located in a rectangular coordinate system A charge of 1 C is at (1 m, 1 m) a charge of 2 C is at (1 m, -1 m) a charge of 3 C is at (-1 m, -1 m) and at (-1 m, 1 m) is a charge of 4 C What is the direction of the net electric field at the coordinate origin due to the four point charges? A) Due north B) Due East C) Due South D) Due west Answer According to Coulomb's Law, the net electric field at a point due to multiple point charges can be calculated by summing the individual electric fields at that point. The electric field at a point is a vector quantity and is defined as the force per unit charge that a positively charged particle would experience if placed at that point. To determine the direction of the net electric field at the coordinate origin, we need to calculate the electric field at that point due to each of the four point charges and then add them together. We can then use the direction of this net electric field vector to determine which option is the correct answer. Let's calculate the electric field at the origin due to each point charge individually: 1) Charge of 1 C at (1 m, 1 m): Using Coulomb's Law, the electric field at the origin due to this charge can be calculated as: E = kq/r^2 Where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the charge (1 C) and r is the distance between the charge and the point of interest (sqrt(2) m). Plugging in the values, we get: E1 = (8.99 x 10^9)(1 C)/(sqrt(2) m)^2 = 9.0 x 10^9 N/C Since this charge is located at the same distance from the origin in both the x and y directions, the electric field vector due to this charge will be in the positive diagonal direction, which can be denoted as NE (northeast). 2) Charge of 2 C at (1 m, -1 m): Using the same equation as before, the electric field at the origin due to this charge can be calculated as: E = (8.99 x 10^9)(2 C)/(2 m)^2 = 2.25 x 10^9 N/C Since this charge is located at a distance of sqrt(2) m in the x direction and sqrt(2) m in the negative y direction, the electric field vector will be in the negative diagonal direction, which can be denoted as SW (southwest). 3) Charge of 3 C at (-1 m, -1 m): Using the same equation as before, the electric field at the origin due to this charge can be calculated as: E = (8.99 x 10^9)(3 C)/(sqrt(2) m)^2 = 1.8 x 10^10 N/C Since this charge is located at a distance of sqrt(2) m in both the x and y directions, the electric field vector will be in the negative diagonal direction, which can be denoted as NW (northwest). 4) Charge of 4 C at (-1 m, 1 m): Using the same equation as before, the electric field at the origin due to this charge can be calculated as: E = (8.99 x 10^9)(4 C)/(2 m)^2 = 4.5 x 10^9 N/C Since this charge is located at a distance of 2 m in the x direction and sqrt(2) m in the positive y direction, the electric field vector will be in the positive diagonal direction, which can be denoted as SE (southeast). Now, we can add these electric field vectors together to find the net electric field at the origin: Net electric field (ENet) = E1 + E2 + E3 + E4 = (9.0 x 10^9 N/C) + (2.25 x 10^9 N/C) + (1.8 x 10^10 N/C) + (4.5 x 10^9 N/C) = 2.67 x 10^10 N/C Since this net electric field vector is in the positive diagonal direction (NE), we can conclude that the answer is A) Due north. Question 4 A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. Calculate the acceleration of the car and the distance it travels in this time. Answer In order to answer this question, we will rely on our knowledge of basic kinematics and the equations of motion. Let's break down the question into two parts - calculating the acceleration of the car and the distance it travels in 10 seconds. Acceleration: The acceleration can be calculated using the equation a = (Vf - Vo)/t, where a is the acceleration, Vf is the final velocity, Vo is the initial velocity and t is the time taken. In this case, the final velocity is 25 m/s, the initial velocity is 0 m/s (since the car starts from rest) and the time taken is 10 seconds. Therefore, the acceleration of the car will be: a = (25 m/s - 0 m/s)/10 s a = 2.5 m/s^2 Therefore, the acceleration of the car is 2.5 m/s^2. Distance: To calculate the distance traveled by the car, we will use the equation d = Vo*t + 0.5*a*t^2, where d is the distance, Vo is the initial velocity, t is the time taken and a is the acceleration. In this case, the initial velocity is 0 m/s, the time taken is 10 seconds and the acceleration is 2.5 m/s^2. Substituting these values, we get: d = (0 m/s)*(10 s) + 0.5*(2.5 m/s^2)*(10 s)^2 d = 0 m + 0.5*25 m = 12.5 m Therefore, the distance traveled by the car in 10 seconds is 12.5 m. Question 5 A parallel-plate capacitor is filled with a dielectric material with a dielectric constant 𝜅 = 4 κ=4. The plates have an area of 1 m 2 1m 2 and are separated by a distance of 0.01 m 0.01m. If a potential difference of 1000 V 1000V is applied across the plates, calculate the capacitance of the capacitor, the charge stored on the plates, and the electric field within the dielectric. Answer The capacitance of a parallel-plate capacitor is calculated by the equation C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space (8.85x10^-12 F/m), A is the area of the plates, and d is the distance between the plates. However, when a dielectric material is introduced between the plates, the effective permittivity becomes ε = ε0𝜅, where 𝜅 is the dielectric constant. Therefore, the capacitance of this parallel-plate capacitor can be calculated as: C = εA/d = ε0𝜅A/d = (8.85x10^-12)(4)(1)/(0.01) = 3.54x10^-9 F To calculate the charge stored on the plates, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Substituting the values we have, the charge stored on the plates is: Q = (3.54x10^-9)(1000) = 3.54x10^-6 C Finally, to determine the electric field within the dielectric, we can use the equation E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the plates. Substituting the values we have, the electric field within the dielectric is: E = (1000)/(0.01) = 100,000 V/m