JEE Main20199 Jan 2019Evening ShiftPhysicsElectrostaticsActual
Two point charges q 1 10 μC and q 2 - 25 μC are placed on the x -axis at x = 1 m and x = 4 m respectively. The electric field in V / m at a point y = 3 m on y -axis is, Take 1 4 π ϵ 0 = 9 × 10 9 N m 2 C - 2
Options
- A- 81   i ^ + 81   j ^ × 10 2
- B81   i ^ - 81   j ^ × 10 2
- C- 63   i ^ + 27   j ^ × 10 2
- D63 i ^ - 27 j ^ × 10 2
Correct answer
D. 63 i ^ - 27 j ^ × 10 2
Step-by-step solution
Electric field due to 10   μC E → 1 = 1 4 π ∈ 0 10 × 10 - 6 r 1 3   r → 1 = 1 4 π ∈ 0 10 × 10 - 6 10 3 - i ^ + 3 j ^ Similarly, electric field due to - 25   μC E → 2 = 1 4 π ∈ 0 25 × 10 - 6 5 3 4 i ^ - 3 j ^ Net electric field, E → = 9 × 10 9 × 10 - 6 - i ^ 10 + 3 j ^ 10 + 4 5 i ^ - 3 5 j ^ = 9 × 10 3 7 i ^ 10 - 3 j ^ 10 = 63 i ^ - 27 j ^ × 10 2   N C