Question

PHY-23102-201910 (UNIVERSITY PHYSICS HOMEWORK SET# 1 DUE DATE: MONDAY, JANUARY 28 Please write neatly and do not forget to put your name Problem 1: Consider two vectors: a-ai + 4, and ธิ-5i-bj. Here i is a unit vector (F|-1) along the z-axis and 3 is a unit vector = 1) along the y-axis. (a) Write down the expressions for vectors a +b and a -0 in terms of the unit vectors i and ) (b) Express vector č- a x 8 in terms of i and j. What is the length of veetor e -č/ia? Problem 2: An imaginary particle has mass m # 5 10-27 kg and charge q +3c 4.8 x 10-19 C. Compare the magnitude of the electric repulsion between two r particles with that of gravitational attraction between them. Problem 3: Two equal charges q1 2-2 x 10-6 C are located at z-0, y-0.3 m and z-0, M0.3 m respectively. What are the magnitude and direction of the total electric force that q and 92 exert on a third charge g-4.0 x 10-6 C located at z 0.1 m, y = 0 m? Problem 4: What is the magnitude of the electric field E at a field point 2.0 m from a point charge q - 4.0 x 109 C? Problem 5: When the terminals of a battery are connected to two parallel conducting plates with a small gap between them, the resulting charges on the plates produce a nearly uniform electrie field E between the plates. Assume that positively charge plane is a lower one and negatively charged p field is known to be lei # 1 x 104 N/C. lane is an upper one, so the feld E is pointing upward. The magnitude of the electric 9.1 × 10-31 kg) is released from (a) If an electron (charge-e H-1.6 × 10-9 C and mass m the rest at the upper plate, what is its acceleration? (b) What speed and kinsetie enengy does it acquire while traveling 1.0 em to the lower plate? Problem 6: Charge Q is umifonnly distributed on a ring of radius R. Find an electric field at a point P on the ring symmetry axis at a distance from its center MacBook Pro

Answer 1

Given a vector = 3i^^+ 4j^^ Vertical-bar j^^ Vertical-bar = 1 b vector = 5i^^- 6j^^ Vertical-bar i^^ Vertical-bar = 1 a vector + b vector = 3i^^+ 4j^^+ 5i^^- 6j^^a vector + b vector = 8i^^- 2j^^a vector - b vector = 3i^^+ 4j^^- 5i^^+ 6j^^a vector - b vector = -2i^^+ 10j^^c = -2i^^+ 10j^^c vector = a vector times b vector = [i^^3 5 j^^4 -6 k^^0 0] = i^^(0 - 0) - j^^(0 - 0) + k^^(-18 - 20) c vector = -38 k^^c vector = c vector/ Vertical-bar c vector Vertical-bar = -38 k^^/38 = -k^^