Question

MCAT Physics Mini-Tutorial: Equipotential Lines & Dipoles

Answer 1

\({\bf{Equipotential~Lines:}}\) dashed lines across which the electric potential of a charge is equal. no work is done when moving charge along an equipotential.
- additional note: work done to move a charge is independent of path

\({\bf{Dipoles:}}\) two equal and opposite sign charges separated by some distance d

electric potential for a dipole:

V = kq/r1 - kq/r2 (negative sign because we add the potential energies but they have opposite signs so the sign in the middle ends up being a subtraction sign)

combining them to have the same denominator would give us

kq(r2-r1)/r1*r2

using the fact that the distances are the same & the fact that (r2-r1) is approx. dcos(theta) for large values of r1 and r2 gives us:

V = kqdcos(theta)/r^2
where theta is the angle between the dipole axis and the reference point

Answer 2

\({\bf{More~Dipole~Equations:}}\)

dipole moment = qd (where q is the magnitude of each charge, d is distance between the charges)

perp. bisector of the dipole: the line that creates a 90 degree angle between the bisector and the dipole axis

E = 1/(4pi * epsilon naught) cross (p / r^3)
note that this is a cross product

torque: using the torque equation Fdsin(theta) on each end of the dipole:

torque = Fsin(theta) * (d/2) + Fsin(theta) * (d/2)

(remember it is d/2 since we are measuring from the point of rotation)

adding them gives us

Fsin(theta) * d

re-writing the electric force as qE gives us

qEsin(theta) * d

now, we know that dipole moment = p = qd so

torque = pEsin(theta)

Answer 3

Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the Third Edition Kaplan Physics & Mathematics Prep Book for the new MCAT

Answer 4

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Answer 13

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