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frank0520
 one year ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r≫s. Immediately after the dipole is released,
what is the magnitude of the force on the dipole?
what is the magnitude of the torque on the dipole?
enter the factor that multiplies 1/ϵ in your answer. Express this factor in terms of q, Q, s, π, and r.
frank0520
 one year ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r≫s. Immediately after the dipole is released, what is the magnitude of the force on the dipole? what is the magnitude of the torque on the dipole? enter the factor that multiplies 1/ϵ in your answer. Express this factor in terms of q, Q, s, π, and r.

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the field from Q is \(\vec E = k\frac{Q}{r^2} \ \hat r\) which given the distances i think you can treat as a uniform parallel field on the dipole of \(E = k\frac{Q}{r^2} \) with \(k = \frac{1}{4 \pi \epsilon_o}\) but the net force on the dipole as a whole should be zero the dipole moment p follows from \(\vec p = q \vec d\) where d is the displacement vector and the torque from \(\vec \tau = \vec p \times \vec E = pE\hat n\) simplifying to \(\tau = pE =q \times s \times E\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\sf \vec F = k\frac{q_1q_2}{r^2}\]Can you use thisssss?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Dangit pressed enter too late >:(

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1The net force by the charge on the dipole cannot be zero since the net force by the dipole on the charge is not. Work out the latter and use Newton's 3rd law.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441978510361:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec F_1 + \vec F_1 = \] \[k\frac{Qq}{r^2} \left[ <\sin \theta, \cos \theta> + <\sin \theta, \cos \theta>\right]\] \[= 2k\frac{Qq}{r^2}\frac{d}{2r}\] \[= \frac{1}{4 \pi \epsilon}Qq\frac{d}{r^3}\] i was operating on the basis \(s << r \implies \ F_{net} = 0\) mea culpa

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441981811457:dw

frank0520
 one year ago
Best ResponseYou've already chosen the best response.0I got this for magnitude of torque: 1. Edipole=(1/4πε)(2p/r3) 2. p=qs 3. F=qE 4. τ=pEsinϑ \[\tau=\frac{ q^2 s^2 }{ 2\pi \epsilon_0 r^3 }\sin (\theta)\] What am I missing?

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1The torque is: \(\tau=k\dfrac{Qqs}{r^2}\)
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