Two charges Q_A = +q and Q_B = - 3q are located on the x-axis at x= 0 and x= d respectively. Where is the electric potential equal to zero?

Question

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The electric potential at a point on the x-axis due to two charges Q_A and Q_B is given by:$$V = k(\frac{Q_A}{r_A} + \frac{Q_B}{r_B})$$where k is the Coulomb constant, r_A is the distance from the point to Q_A, and r_B is the distance from the point to Q_B.If the electric potential is equal to zero, then:$$k(\frac{Q_A}{r_A} + \frac{Q_B}{r_B}) = 0$$$$k(\frac{q}{x} + \frac{-3q}{x-d}) = 0$$$$\frac{q}{x} = \frac{3q}{x-d}$$$$x = \frac{d}{4}$$

Two Charges QA = +Q and QB = - 3q

Two Charges Q_A = +Q and Q_B = - 3q