MCAT Physics Mini-Tutorial: Equipotential Lines & Dipoles

9 months ago\({\bf{Equipotential~Lines:}}\) 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

9 months ago\({\bf{More~Dipole~Equations:}}\) 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)

9 months agoAnyway, 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

9 months ago