A point charge Q -485 nC and two unknown point charges % and g2 are placed as sh

Question

A point charge Q -485 nC and two unknown point charges % and g2 are placed as shown in the fi re The electric field at the origin O due to charges QNa and 92 is equal to 832 N/C directed at an angle 29 from the negative y-axis in the fourth quadrant. The distances of charges from the origin are r2.13 m, 1 1.29 m, and ry 1.66 m. On your solution paper 1) Use E. E, EQ. E as notations for the electric fields of charges 1) draw and label the electric field vectors Eo and E at the origin . Q . and net field respectively 2) Show the solution of the questions below neatly, explaining your work step by step. The solution starts with a physics lawlequation used in terms of the symbols shown, then numerical substitutions, and final answer boxed 91 30° O 292 Part A What is the magnitude and the sign of charge qi?Explain the choice of the sign of the charge nC Submit My Answers Give Up

Explanation / Answer

a] Y component of field, Ey = - kQ/r^2 sin 30 degree - kq1/r1^2

-832 cos 29 degree = -9e9*-485e-9/2.13^2 *0.5 - 9e9*q1/1.29^2

q1 =  [-832 cos 29 degree +9e9*-485e-9/2.13^2 *0.5]/[-9e9/1.29^2] = 2.235*10^-7 C

= 223.5 nC

It has to be positive because if negative, it will make positive y directionof net electric field which should be not.

b] x component of field, Ex = kQ/r^2 cos 30 degree - kq2/r2^2

832 sin 29 degree = 9e9*-485e-9/2.13^2 *0.866 - 9e9*q2/1.66^2

q2 =  [832 sin 29 degree -9e9*-485e-9/2.13^2 *0.866]/[-9e9/1.66^2] = -3.786*10^-7 C

= -378.6 nC

It has to be negative because if positive, it will make x direction of net electric field negative which should not be.

c] V = summation kq/r. = 9e9*[-485e-9/2.13 + 223.5e-9/1.29 - 378.6/1.66]

= -2543 V

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