A charge +q is at the origin. A charge -2q is at x = 6.20 m on the +x axis. (a)

Question

A charge +q is at the origin. A charge -2q is at x = 6.20 m on the +x axis.
(a) For what finite value of x is the electric field zero?
........m

(b) For what finite values of x is the electric potential zero? (Note: Assume a reference level of potential V = 0 at r = .)
Smallest value of x:
........m

Largest value of x:
........m

Explanation / Answer

E = kq/r^2 let the distance from q be x. this means distance from -2q = x+6.2 now kq/x^2 - k2q/(x+6.2)^2 = 0 => (x+6.2)^2 = 2x^2 => x^2 -12.4x - 38.44 = 0 x = -2.5 and x = 14.96 form origin and q b) potential = kq/r for potential zero we have the point b/w the two charges kq/y = k2q/(6.2-y) => 2y =- y+6.2 => y = 6.2/3 = 2.06 form origin and q

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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