Electric field due to q2 at point p1 is (-1667.0) i + (18340.0) j Electric field due to q3 at point p1 is (-2496.0) i + (–1628.0) j Electric field due to q1 at point p2 is (-6311.0) i + (2470.0) j Electric field due to q2 at point p2 is (-11585.0) î + (–6951.0) ĵ Electric field due to q3 at point p2 is (4657.0) i + (11310.0) j c) Find the net Electric field at p1 .

Electric field due to q2 at point p1 is (-1667.0) i + (18340.0) j Electric field due to q3 at point p1 is (-2496.0) i + (–1628.0) j Electric field due to q1 at point p2 is (-6311.0) i + (2470.0) j Electric field due to q2 at point p2 is (-11585.0) î + (–6951.0) ĵ Electric field due to q3 at point p2 is (4657.0) i + (11310.0) j c) Find the net Electric field at p1 .

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Please try to answer the question within 25-30 minutes
I need the answer as soon as possible
The figure shows an electric dipole. What is the magnitude of the dipole's electric field at point P? Assume that q = 7.88 × 10-6 C, d = 2.07 x 10-6 m, and r = 9.46 cm. +g d/2 d/2 -4 Number i Units N/C or V/m
In mechanical resistance
1. please write clean and as much detail as you can 2. please draw the FBD since I am not understanding the temperature placement on this question 3. to make your job easier the following formula apply : Q_cond=Q_conv+Q_radi KA(T_i-T_o )/L=hA(T_s-T_∞ )+ε σ A(T_s^4-T_∞^4 )
only first attempt please
A single square loop of wire is placed in the center of a large solenoid as shown. The solenoid has a radius 5.00 cm and the square loop is 3.0 cm on a side. The solenoid is 0.20 cm long and wound with 200 turns of wire. If the current in the solenoid is 3.00 A, what is the flux through the square loop? If the current in the solenoid drops to zero in 0.500 seconds, what is the magnitude of the average induced emf in the square loop?
A uniform electric field of magnitude 260 V/m is directed in the positive x direction. A +15.0 μC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge field system? ] (b) Through what potential difference does the charge move? v
Q2] (10 Marks) Find the relation between backwards finite difference and average operator.
Find the expression of Hooke's Law using indicial notations: 1 1+v E11 = [011 − v(022 + 033)]; &12 = 7 E 22 = E33 = 1 [022 - V(0₁1 + σ33)]; E23 = 1 ;[033 − v(011 + 022)]; &13 = 1 + v E - 012 E - 023 1+ v - 013
Numerical analysis