A point charge is fixed at and a point charge at in vacuum

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electrostatic fields and potential · Electrostatic fields and potential

Electricity, Magnetism and Electromagnetism, Ch Electric charge, Coulomb's law and the electric field

Problem

A point charge +q+q is fixed at (0,0,+d/2)(0,0,+d/2) and a point charge −q-q at (0,0,−d/2)(0,0,-d/2) in vacuum, with q>0q > 0; the dipole moment has magnitude p=qdp = qd and points along +z+z (from the negative to the positive charge). Point PP lies on the perpendicular bisector, at (x,0,0)(x,0,0) with x>0x > 0. ((a)) By resolving the two point-charge fields into components, show that the field at PP is E=qd4πε0(x2+d2/4)3/2E = \frac{qd}{4\pi\varepsilon_0\left(x^{2} + d^{2}/4\right)^{3/2}} directed opposite to p\mathbf{p} (along −z-z), the components along xx cancelling. ((b)) Show that for x≫dx \gg d this becomes E≈p/4πε0x3E \approx p/4\pi\varepsilon_0 x^{3}, and hence find the ratio of the far field on the bisector to the far axial field 2p/4πε0r32p/4\pi\varepsilon_0 r^{3} at the same distance from the dipole's centre. ((c)) A molecule has p=3.4×10−30 C mp = 3.4\times10^{-30}\ \mathrm{C\,m}, and its charge separation is much smaller than 2.0 nm2.0\ \mathrm{nm}. Find the magnitude of its field at a point on its perpendicular bisector 2.0 nm2.0\ \mathrm{nm} from its centre, and state the direction.

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