A positive charge of magnitude Q_1 = 85 nC is located at the origin. A negative
ID: 1602681 • Letter: A
Question
A positive charge of magnitude Q_1 = 85 nC is located at the origin. A negative charge Q_2 = - 65 nC is located on the positive x-axis at x = 145 cm from the origin. The point P is located y = 17 cm above charge Q_2. Calculate the x-component of the electric field at point P due to charge Q_1. Write your answer in units of N/C. E_x, 1 = Calculate the y-component of the electric field at point P due to charge Q_1. Write your answer in units of N/C. Calculate the y-component of the electric field at point P due to the Charge Q_2. Write your answer in units of N/C. Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C. Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C. Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.Explanation / Answer
Given
Q1 = 8.5 nC at origin and
Q2 = -6.5 nC at (14.5 cm,0 cm)
point P is at (14.5 cm, 17 cm)
the angle theta is tan theta = (0.17/0.145) ==> theta = 49.54 degrees
the electric field is E = kq/r^2
the distance from the charge Q1 to point P is r1 = sqrt(0.145^2+0.17^2) m = 0.2234 m
a) Ex1 = kQ1cos theta/r1^2
Ex1 = 9*10^9*8.5*10^-9*cos 49.54/(0.2234)^2 N/C = 994.6820 N/C
b)
Ey1 = 9*10^9*8.5*10^-9*sin 49.54/(0.2234)^2 N/C = 1166.271 N/C
c)
Ey2 = kQ2 sin theta/r2^2
= 9*10^9*(-6.5*10^-9) sin 90 /0.17^2 N/c
= -2024.2215 N/C
d) y component of electric field at point P is
Ey = Ey1 + Ey2 = 1166.271 -2024.2215 N/C = -857.9505 N/C
e) magnitude of the electric field at point P is
E = Ex+ Ey
= 994.6820 i +(-857.9505) j
E = sqrt(Ex^2+Ey^2)
= sqrt(994.6820^2+(-857.9505)^2) N/C
= 1313.572 N/C
f) angle theta = arc tan ((-857.9505)/994.6820)==> theta = -40.77 degrees
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.