The electric field at the point x = 10.0 cm and y = 0 points in the positive x d

Question

The electric field at the point x = 10.0 cm and y = 0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x = 25.0 cm and y = 0 the electric field points in the positive x direction with a magnitude of 30.0 N/C. Assume that this electric field is produced by a single, point charge.

(a) Find the location of the point charge.
(x, y) = ( , ) cm

(b) Find the sign and magnitude of its charge.

Explanation / Answer

Since magnitude is more at x=25 the charge should be closer to x=25 and let it be at x=d, SO now 10 = Kq/(d-0.1)^2 and 30 = Kq/(d-0.25)^2 Dividing both we get 3 = (d-0.1)^2/(d-0.25)^2 SOlving for d, we get d = (25sqrt(3) - 1)/(10sqrt(3)-10) = 2.591m ANswer -> (x,y)=(d,0) Putting this in one of the equations, you get 'q'. (Sorry but i dont have a calculator right now :( )

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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