COMEDK20269 May 2026Morning ShiftPhysicsElectrostaticsActual
Point charge 2 C, 2 C , and -2C are placed at the three vertices of a right-angled triangle in air. [as shown in the figure below] What is the electric field at a point P on the hypotenuse that is equidistant from all three charges. Given distances XP = YP = ZP = 0.5 m
Options
- A7 2 10^9 NC⁻¹ along PY
- B7 2 10¹⁰ NC⁻¹ along PY
- C0 72 10^9 NC⁻¹ along YP
- D0 72 10¹⁰ NC⁻¹ along YP
Correct answer
B. 7 2 10¹⁰ NC⁻¹ along PY
Step-by-step solution
Let the charges at vertices X , Y , and Z be q_X = 2 C , q_Y = -2 C , and q_Z = 2 C respectively. Point P is on the hypotenuse XZ and is equidistant from X and Z , meaning it is the midpoint of XZ . The electric field at P due to the charge at X is directed away from X (along PZ ) with magnitude: E_X = 1 4 ₀ q_X (XP)^2 The electric field at P due to the charge at Z is directed away from Z (along PX ) with magnitude: E_Z = 1 4 ₀ q_Z (ZP)^2 Since q_X = q_Z = 2 C and XP = ZP = 0.5 m , the magnitudes are equal: E_X = E