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.
ScanSolve AI Answer
Step 1: Identify the given values and convert units to SI units. The charges are: q_1 = 2.1 × 10^-5 C q_2 = 4q_1 = 4 × (2.1 × 10^-5 C) = 8.4 × 10^-5 C The distances are: Distance between q_1 and q_2 is d = 10 cm = 0.10 m. Distance from q_1 to the point P is r_1 = 8 cm = 0.08 m. Distance from q_2 to the point P is r_2 = 6 cm = 0.06 m. The electrostatic constant is k = 9 × 10^9 N m^2/C^2. Step 2: Determine the geometry of the charges and the point P. We check if the point P forms a right-angled triangle with the charges q_1 and q_2. r_1^2 + r_2^2 = (0.08 m)^2 + (0.06 m)^2 = 0.0064 m^2 + 0.0036 m^2 = 0.0100 m^2. d^2 = (0.10 m)^2 = 0.0100 m^2. Since r_1^2 + r_2^2 = d^2, the point P forms a right-angled triangle with q_1 and q_2, with the right angle at point P. This means the electric field vectors E_1 and E_2 at point P will be perpendicular to each other. Step 3: Calculate the electric field intensity due to each charge. The formula for the electric field intensity due to a point charge is E = (k|q|)/(r^2). For charge q_1: E_1 = (k q_1)/(r_1^2) E_1 = (9 × 10^9 N m^2/C^2) × (2.1 × 10^-5 C)(0.08 m)^2 E_1 = 18.9 × 10^4 N m^2/C0.0064 m^2 E_1 = 29531250 N/C = 2.953125 × 10^7 N/C For charge q_2: E_2 = (k q_2)/(r_2^2) E_2 = (9 × 10^9 N m^2/C^2) × (8.4 × 10^-5 C)(0.06 m)^2 E_2 = 75.6 × 10^4 N m^2/C0.0036 m^2 E_2 = 210000000 N/C = 2.1 × 10^8 N/C Step 4: Calculate the resultant electric field intensity. Since $