Dimensions Trick: ε₀ dΦₑ/dt in 60 Seconds! 🔥 | JEE Physics

❓ Question

If $\varepsilon_0$ denotes the permittivity of free space and $\phi_E$ is the electric flux through the area bounded by a closed surface, then the dimensions of

$${\epsilon }_0\frac{d{\varphi }_E}{dt}$$

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:

$${\varphi }_E=\oint \vec{E}\cdot d\vec{A}$$

So dimensionally:

$$[{\varphi }_E]=[E][A]$$

---

🔹 Step 2 — Dimensions of electric field

Electric field:

$$E=\frac{F}{q}$$

Where:

• Force $F$ has dimensions $[MLT^{-2}]$
  

• Charge $q$ has dimensions $[IT]$
  

So:

$$[E]=\frac{ML{T}^{-2}}{IT}=ML{T}^{-3}{I}^{-1}$$

---

🔹 Step 3 — Dimensions of electric flux

Area:

$$[A]=L^2$$

Hence:

$$[{\varphi }_E]=(ML{T}^{-3}{I}^{-1})(L^2)=M{L}^3{T}^{-3}{I}^{-1}$$

---

🔹 Step 4 — Take time derivative of flux

$$\frac{d{\varphi }_E}{dt}\Rightarrow [M{L}^3{T}^{-4}{I}^{-1}]$$

---

🔹 Step 5 — Dimensions of permittivity of free space

From Coulomb’s law:

$$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$$

So:

$$[{\epsilon }_0]=\frac{q^2}{F{r}^2}=\frac{(IT{)}^2}{(ML{T}^{-2})(L^2)}=M^{-1}{L}^{-3}{T}^4{I}^2$$

---

🔹 Step 6 — Multiply $\varepsilon_0$ with $\dfrac{d\phi_E}{dt}$

$$[\varepsilon_0]\left[\frac{d\phi_E}{dt}\right] = (M^{-1}L^{-3}T^{4}I^{2})(ML^3T^{-4}I^{-1})$$

Cancel terms:

$$=I$$

---

✅ Final Answer

$$\boxed{\text{Electric current}}$$

---

🔥 Important JEE Insight

This expression appears directly in Maxwell’s equations:

$${\epsilon }_0\frac{d{\varphi }_E}{dt}\text{has dimensions of current}$$

That’s why it represents displacement current in electromagnetism.

📌 JEE often checks:

• Concept of flux

• Dimensional consistency

• Link with displacement current