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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Step 1: Define the charges and their separation. Let the first charge be Q_1 = +9q and the second charge be Q_2 = -q. The distance between the charges is d = 10 cm. The electric potential V due to a point charge Q at a distance r is given by V = (kQ)/(r), where k is Coulomb's constant. Step 2: Set up the equation for zero electric potential. We are looking for a point where the total electric potential is zero. Let this point be P. The total potential at P is the sum of the potentials due to Q_1 and Q_2: V_P = V_1 + V_2 = (kQ_1)/(r_1) + (kQ_2)/(r_2) where r_1 is the distance from Q_1 to P, and r_2 is the distance from Q_2 to P. We want V_P = 0, so: (k(+9q))/(r_1) + (k(-q))/(r_2) = 0 (9q)/(r_1) = (q)/(r_2) (9)/(r_1) = (1)/(r_2) 9r_2 = r_1 Step 3: Analyze the possible locations for point P. There are two regions on the line joining the charges where the potential can be zero: Region A: Between the charges. Let the point P be at a distance x from Q_1 (+9q). Then r_1 = x. The distance from Q_2 (-q) will be r_2 = d - x = 10 - x. Substitute these into the equation 9r_2 = r_1: 9(10 - x) = x 90 - 9x = x 90 = 10x x = 9 cm This means the point is 9 cm from +9q. This is a valid solution as 0 < 9 < 10. Region B: Outside the charges, to the right of -q. Let the point P be at a distance x from Q_1 (+9q). Then r_1 = x. The distance from Q_2 (-q) will be r_2 = x - d = x - 10. Substitute these into the equation 9r_2 = r_1: 9(x - 10) = x 9x - 90 = x 8x = 90 x = (90)/(8) = (45)/(4) = 11.25 cm This means the point is 11.25 cm from +9q. This is also a valid solution as 11.25 > 10. Region C: Outside the charges, to the left of +9q. Let the point P be at a distance x' to the left of Q_1 (+9q). Then r_1 = x'. The distance from Q_2 (-q) will be r_2 = d + x' = 10 + x'. Substitute these into the equation 9r_2 = r_1: 9(10 + x') = x' 90 + 9x' = x' 90 = -8x' x' = -(90)/(8) = -11.25 cm Since distance cannot be negative, there is no solution in this region. This is expected because the zero potential point must be closer to the charge with the smaller magnitude, which is -q. Step 4: Compare the solutions with the given options. We found two points where the electric potential is zero: 1. 9 cm from +9q. 2. 11.25 cm from +9q (which is 1.25 cm from -q). Let's check the options: (1) 1 cm from +9q (Incorrect) (2) 9 cm from -q. If this is between the charges, it means 1 cm from +9q. (Incorrect) (3) 9 cm from +9q (Correct, matches our first solution) (4) 5 cm from -q. If this is between the charges, it means 5 cm from +9q. (Incorrect) ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._