AP EAMCET201921 Apr 2019Morning ShiftPhysicsElectrostaticsActual
The electric field intensity at a point on the axis of an electric dipole in air is 4 NC ⁻¹ . Then the electric field intensity at a point on the equatorial line which is at a distance equal to twice the distance on the axial line and if the dipole is in a medium of dielectric constant 4 is
Options
- A1 NC ⁻¹
- B1 8 NC ⁻¹
- C16 NC ⁻¹
- D1 16 NC ⁻¹
Correct answer
D. 1 16 NC ⁻¹
Step-by-step solution
As, electric field intensity on the axis, aligned E_ axis = 2 k P r^3 4= 2 k P r^3 for r & >a & ( given, E_ axis =4 ) aligned Electric field intensity on equatorial line, E_ eq = k^ P r₁^3 where, r₁=2 r and k^ = k 4 So, E_ eq = k P 4 8 r^3 Now, from the Eq. (i), we get E_ eq = 2 4 8 = 1 16 NC ⁻¹ Hence, the correct option is (d).