2 positive charges, Q and q are held at rest and separated by a distance of d. C

Question

2 positive charges, Q and q are held at rest and separated by a distance of d. Charge Q is larger than charge q. The particle with charge q has a mass of m.

a) What is the magnitude and direction of the electric field midway between the two charges?
b) What is the potential midway between the charges?
c) The particle with charge q moves off the the right when released from rest. The particle with charge Q is held fixed in position. What speed with it ultimately achieve (Assume that non-relativistic physics applies)?

Explanation / Answer

a) According to Coulomb's Law, E = kq/r^2. So midway between the charges, E = kQ/(d/2)^2 - kq/(d/2)^2 = 4k(Q-q) / d^2 Since Q is larger than q, the direction is from Q to q. b) We have V = kq/r. So midway between the charges, V = kQ/(d/2) - kq/(d/2) = 2k(Q-q) / d c) The initial potential energy of q is: U = kQq/d And we have U = K. So kQq/d = 1/2mv^2. Thus, v = sqrt(2kQq/dm). Hope it's clear.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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