Past exam of the physics 2 paper of the University of Oxford / 2021 / 8

Take the grounded plane as $z=0$. The method of images replaces it, in $z>0$, by an image charge $-q$ at $z=-d$ and an image charge $+q$ at $z=-2d$. These make the electric potential zero on the plane; the uniqueness theorem for the electrostatic potential establishes the solution in the physical half-space.

At $z=3d/2$, both real point charges contribute upward electric fields, each of magnitude $kq/(d/2{)}^2$. The nearer image charge contributes downward and the farther one upward. HenceThis is upward for $q>0$; retaining signed $q$ automatically handles the opposite sign. No self-electric field is involved because the evaluation point is distinct from every point charge.

For image charges in a sixty-degree conducting wedge, use polar coordinates in a cross-section normal to the planes' intersection, putting the physical wedge at $0<\theta <\pi /3$ and the real point charge at $(r,\theta )=(d,\pi /6)$. Repeated reflection in the two grounded faces produces five image charges, all at radius $d$ and the same coordinate along the intersection:Equivalently, the positive copies are at $\theta =\pi /6+2k\pi /3$ and the negative copies at $\theta =-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the original point charge lies inside the physical wedge.

For any observation point on either face, reflection pairs every positive point charge with an equidistant negative one. Their Coulomb potentials cancel, so both grounded faces have zero electric potential. The electric potential also tends to zero at infinity. The uniqueness theorem for the electrostatic potential then identifies this method of images construction with the physical solution.

Let ${\hat{\mathbf{e}}}_r$ point from the intersection to the real point charge. Reflection symmetry cancels the component of electrostatic force perpendicular to the wedge bisector. An image charge $q_j$ at angular separation $\Delta$ from the real point charge gives radial forceThe two adjacent negative image charges have $\Delta =\pm 60^{\circ }$, distance $d$, and combined radial force $-k{q}^2/d^2$. The two positive image charges have $\Delta =\pm 120^{\circ }$, distance $\sqrt{3}d$, and combined radial force $+k{q}^2/(\sqrt{3}{d}^2)$. The opposite negative image charge is at distance $2d$ and contributes $-k{q}^2/(4d^2)$. ThusThe electrostatic force is towards the intersection, independently of the sign of $q$. One sums the five image charge electric fields and multiplies by $q$; the factor $1/2$ used in induced-charge electrostatic energy must not be inserted into this direct force calculation.