A point charge q1 = -2.00 nC is at the point x = -0.800 m, y = -0.600 m, and a s

Question

A point charge q1 = -2.00 nC is at the point x = -0.800 m, y = -0.600 m, and a second point charge q2 = +3.00 nC is at the point x = 0.300 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges. A point charge q1 = -2.00 nC is at the point x = -0.800 m, y = -0.600 m, and a second point charge q2 = +3.00 nC is at the point x = 0.300 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges. A point charge q1 = -2.00 nC is at the point x = -0.800 m, y = -0.600 m, and a second point charge q2 = +3.00 nC is at the point x = 0.300 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Explanation / Answer

E1 = k*q1/r1^2

= 9*10^9*2*10^-9/(0.8^2 + 0.6^2)

= 18 N/c

let theta is the angle made by E1 with -x axis

so, theta = tan^-1(0.6/0.8)

= 36.87 degrees below +x axis

so,

E1x = -E1*cos(36.87)

= -18*cos(36.87)

= -14.4 N/c

E1y = -E1*sin(36.87)

= -18*sin(36.87)

= -10.8 N/c

E2 = k*q2/r2^2

= 9*10^-9*3*10^-9/0.3^2

= 300 N/c

E2x = -E2

= -300 N/c

E2y = 0

so, Enetx = E1x + E2x

= -14.4 - 300

= -314.4 N/c

Enety = E1y + E2y

= -10.8 + 0

= -10.8 N/c

Enet = sqrt(Enetx^2 + Enety^2)

= sqrt(314.4^2 + 10.8^2)

= 314.6 N/c

direction : theta = tan^-1(Enety/Enetx)

= tan^-1(10.8/314.4)

= 1.97 degrees below -x axis

or 181.97 degrees with +x axis in counter-clockwise direction.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.