Question

Physics 133 Studio, Winter 2019 Turn-In Homework 2: Electric Field of Discrete Charge Distributions The solutions to the problems in this homework are algebraic expressions in terms of given variables (a, d, q, ) and constants (K, eo, π' ). On the second page of this handout you will find suggestions for writing algebraic solutions Problem 1 Three point charges are fixed in the x-y plane: 1+35q is at (4d, 3d) q2169q is at (12d,-5d) 3+270q is at (-9d,-12d) The goal is to find the net electric field at the origin, (0, 0) m. 1. Sketch the situation. 2. Consider the electric field due to charge q1 at the origin. On your drawing for Question (1), sketch and label the electric field vector at the origin due to this charge. (Recommendation: use a colored pen or pencil.) a. Find the magnitude of the electric field due to this charge. Show your work. Find the x- and y-components of the electric field due to this charge. Show your work. (Solution: E1,--28d2; Ely b. c. 28Kq 21Kq 3. Repeat step 2 for charges q2 and q3. (In your drawing, use a different color for each field vector.) 4. Find the net electric field at the origin. Show your work. 5. Sketch the net electric field vector at the origin Problem 2 Consider the arrangement of charges shown in the figure. Q > 0 1. Find the components of the net electric field +70 +50 at the center of the rectangle. Explain and show your work, including any assumptions and simplifications that you make. As part of your solution, sketch the electric field due to each charge. o Hint: stop and take time to think before -110 -110 deciding how to solve this part. 2. What is the magnitude of the net electric 2 field? Show your work +50 -70

Answer 1

problem 2 :

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using Pythagorean theorem

AD = BE = sqrt(a² + b²)

r_A = distance of charge at A from O = r_B = r_D = r_E = sqrt(a² + b²)/2

r_C = r_F = a/2

E_A = electric field by charge at A = k Q_A /r_A² = k (7Q)/(sqrt(a² + b²)/2)² = 28 kQ/ (a² + b²)

E_B = electric field by charge at B = k Q_B /r_B² = k (5Q)/(sqrt(a² + b²)/2)² = 20 kQ/ (a² + b²)

E_D = electric field by charge at D = k Q_D /r_D² = k (7Q)/(sqrt(a² + b²)/2)² = 28 kQ/ (a² + b²)

E_E = electric field by charge at E = k Q_E /r_E² = k (5Q)/(sqrt(a² + b²)/2)² = 20 kQ/ (a² + b²)

E_C = electric field by charge at C = k Q_C /r_C² = k (11Q)/(a/2)² = 44 kQ/ a²

E_F = electric field by charge at F = k Q_F /r_F² = k (11Q)/(a/2)² = 44 kQ/ a²

E_C and E_F being equal and opposite cancel out.

E_B and E_E being equal and opposite cancel out.

hence net electric field at O is given as

E = E_D + E_A = 28 kQ/ (a² + b²) + 28 kQ/ (a² + b²) = 56 kQ/ (a² + b²)