Dimensions Trick: ε₀ dΦₑ/dt in 60 Seconds! 🔥 | JEE Physics
❓ Question If ε 0 \varepsilon_0 denotes the permittivity of free space and ϕ E \phi_E is the electric flux through the area bounded by a closed surface, then the dimensions of ε 0 d ϕ E d t are the same as that of what physical quantity? 🖼️ Question Image ✍️ Short Solution This is a pure dimensional-analysis + concept question from Electrostatics + Maxwell’s equations . No numbers, no calculus — just definitions and dimensions . 🔹 Step 1 — Meaning of electric flux Electric flux through a closed surface is: ϕ E = ∮ E ⃗ ⋅ d A ⃗ So dimensionally: [ ϕ E ] = [ E ] [ A ] 🔹 Step 2 — Dimensions of electric field Electric field: E = F q Where: Force F F has dimensions [ M L T − 2 ] [MLT^{-2}] Charge q q q has dimensions [ I T ] [IT] So: [ E ] = M L T − 2 I T = M L T − 3 I − 1 🔹 Step 3 — Dimensions of electric flux Area: [ A ] = L 2 Hence: [ ϕ E ] = ( M L T − 3 I − 1 ) ( L 2 ) = M L 3 T − 3 I − 1 🔹 Step 4 — Take time derivative of flux ...