Final answer:
To find the expression for x0, we consider the potential due to two charges, q1 and q2, at positions x1 and x2. The potential expression is set to zero at infinity, and taking the limit as x approaches infinity leads to the condition that x1 and x2 should approach negative infinity. Given that -3 m < x0 < 1 m, the expression for x0 in terms of q1, q2, x1, and x2 is x0 = (x1 + x2) / 2.
Step-by-step explanation:
To find the expression for x0, we need to consider the potential due to two charges, q1 and q2, at positions x1 and x2 respectively. The expression for the potential at a point (x, 0) is given by:
V = k * (q1 / (x - x1)) + k * (q2 / (x - x2)),
where k is the electrostatic constant. We are given that the potential vanishes at infinity, so let's take the limit of V as x approaches infinity. In the limit, the potential should approach zero, which means the terms in the expression for V should approach zero. Setting this up, we have:
lim(x→∞) V = lim(x→∞) k * (q1 / (x - x1)) + k * (q2 / (x - x2)) = 0.
Since the potential at infinity is chosen to be zero, the terms with q1 and q2 should approach zero. This implies that the denominators in each term should approach zero as well. Setting up this condition, we get:
lim(x→∞) (x - x1) = ∞ and lim(x→∞) (x - x2) = ∞.
This means that x1 and x2 should both be less than infinity, but greater than negative infinity. From the given information, we know that -3 m < x0 (the x-coordinate of the point where potential is zero) < 1 m. Therefore, the expression for x0 in terms of q1, q2, x1, and x2 is:
x0 = (x1 + x2) / 2.