 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.