JEE Main20248 Apr 2024Evening ShiftPhysicsUnits and DimensionsActual
If ₀ is the permittivity of free space and E is the electric field, then ₀ E ^2 has the dimensions :
Options
- A[ M ⁻¹ ~L ⁻³ ~T ^4 ~A ^2 ]
- B[ M L ^2 ~T ⁻² ]
- C[ M ^ L ⁻² TA ]
- D[ M L ⁻¹ ~T ⁻² ]
Correct answer
D. [ M L ⁻¹ ~T ⁻² ]
Step-by-step solution
aligned & E = KQ R ^2 & E = Q 4 ₀ R ^2 & ₀= Q 4 R ^2 E & Now, ₀ E ^2= Q 4 R ^2 E E ^2= Q 4 R ^2 E & [ ₀ E ^2 ]= [ QE R ^2 ]= [ Q ][ E ] [ R ^2 ] = [ Q ] [ R ^2 ] [ W ] [ Q ][ R ] & = [ W ] [ R ^3 ] = ML ^2 ~T ⁻² ~L ^3 = ML ⁻¹ ~T ⁻² aligned