This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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To find the point where the electric potential is zero, we use the formula for the electric potential due to a point charge: V = (kq)/(r) where k is Coulomb's constant, q is the charge, and r is the distance from the charge. For two point charges, the total electric potential at a point is the algebraic sum of the potentials due to each charge: V_total = V_1 + V_2 = (kq_1)/(r_1) + (kq_2)/(r_2) We want to find the point where V_total = 0. Given: q_1 = +4 = +4 × 10^-6 C q_2 = -2 = -2 × 10^-6 C Distance between charges, d = 30 cm Let's consider a point on the line joining the charges. Since the charges have opposite signs, there are two regions where the potential can be zero: 1. Between the charges. 2. Outside the charges, on the side of the smaller magnitude charge (i.e., to the right of q_2). Case 1: The point is between the charges. Let the point be at a distance x from q_1. Then the distance from q_2 will be r_2 = d - x = (30 - x) cm. Setting the total potential to zero: (kq_1)/(x) + (kq_2)/(d-x) = 0 Since k is a non-zero constant, we can divide by k: (q_1)/(x) + (q_2)/(d-x) = 0 (q_1)/(x) = -(q_2)/(d-x) Substitute the values for q_1 and q_2: +4 × 10^-6 Cx = --2 × 10^-6 C30 cm - x (4)/(x) = (2)/(30 - x) Now, solve for x: 4(30 - x) = 2x 120 - 4x = 2x 120 = 6x x = (120)/(6) x = 20 cm This means the electric potential is zero at a point 20 cm from the +4 charge. This point is between the two charges. Case 2: The point is outside the charges, to the right of q_2. Let the point be at a distance x from q_1. Then the distance from q_2 will be r_2 = x - d = (x - 30) cm. Setting the total potential to zero: (kq_1)/(x) + (kq_2)/(x-d) = 0 (q_1)/(x) = -(q_2)/(x-d) +4 × 10^-6 Cx = --2 × 10^-6 Cx - 30 cm (4)/(x) = (2)/(x - 30) 4(x - 30) = 2x 4x - 120 = 2x 2x = 120 x = 60 cm This means the electric potential is also zero at a point 60 cm from the +4 charge (which is 30 cm to the right of the -2 charge). Comparing our results with the given options: (A) 10 cm from +4µC (B) 20 cm from +4µC (C) 15 cm from +4µC (D) 25 cm from [partially visible] The value x = 20 cm from +4 matches option (B). The final answer is B.