The dimensional formula of 1/2 ε₀ E² (ε₀ = permittivity of vacuum and E= electric field) is M^a L^b T^c. The value of 2a-b+c= _______.

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q26 – JEE-MAIN 2026

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Home Physics Units & MeasurementsQuestion #4

The dimensional formula of

\frac{1}{2} ϵ_{0} E^{2}

(

ϵ_{0} =

permittivity of vacuum and

E =

electric field) is

M^{a} L^{b} T^{c}

. The value of

2 a - b + c =

_______.

Clue 1 of 3: Understand the physical quantity

The expression $\frac{1}{2} ϵ_{0} E^{2}$ represents the energy density of an electric field.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the dimensions of

ϵ_{0}

and

E

✦ Active

The dimensions of permittivity of vacuum ( $ϵ_{0}$ ) can be derived from Coulomb's Law, $F = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}}$ . Rearranging for $ϵ_{0}$ , we get $[ϵ_{0}] = \frac{[q]^{2}}{[F] [r]^{2}}$ . Using $[q] = [I T]$ , $[F] = [M L T^{- 2}]$ , and $[r] = [L]$ :

[ϵ_{0}] = \frac{[I T]^{2}}{[M L T^{- 2}] [L]^{2}} = [M^{- 1} L^{- 3} T^{4} I^{2}]

The dimensions of the electric field ( $E$ ) can be derived from the force on a charge, $F = q E$ . Rearranging for $E$ , we get $[E] = \frac{[F]}{[q]}$ :

[E] = \frac{[M L T^{- 2}]}{[I T]} = [M L T^{- 3} I^{- 1}]

Step 2: Calculate the dimensional formula of

\frac{1}{2} ϵ_{0} E^{2}

○ Expand

The constant $\frac{1}{2}$ is dimensionless. Now, we combine the dimensions of $ϵ_{0}$ and $E^{2}$ :

[ϵ_{0} E^{2}] = [M^{- 1} L^{- 3} T^{4} I^{2}] \times ([M L T^{- 3} I^{- 1}])^{2}

Simplifying the exponents:

[ϵ_{0} E^{2}] = [M^{- 1} L^{- 3} T^{4} I^{2}] \times [M^{2} L^{2} T^{- 6} I^{- 2}]

[ϵ_{0} E^{2}] = [M^{(- 1 + 2)} L^{(- 3 + 2)} T^{(4 - 6)} I^{(2 - 2)}] = [M^{1} L^{- 1} T^{- 2}]

This dimensional formula corresponds to energy density (Energy/Volume), which is $[M L^{2} T^{- 2}] / [L^{3}] = [M L^{- 1} T^{- 2}]$ . This confirms our calculation.

💡 Teacher's Secret Hint

Remember that constants like $\frac{1}{2}$ or $4 π$ are dimensionless and do not affect the dimensional formula.

Step 3: Compare with

M^{a} L^{b} T^{c}

and calculate

2 a - b + c

○ Expand

Comparing the derived dimensional formula $[M^{1} L^{- 1} T^{- 2}]$ with the given form $M^{a} L^{b} T^{c}$ , we identify the exponents:

a = 1, b = - 1, c = - 2

Now, we calculate the value of $2 a - b + c$ :

2 a - b + c = 2 (1) - (- 1) + (- 2)

2 a - b + c = 2 + 1 - 2 = 1

💡 Teacher's Secret Hint

Pay close attention to the signs when substituting the values of $a$ , $b$ , and $c$ into the final expression.

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Physics Question 26 – JEE-MAIN 2026

Clue 1 of 3: Understand the physical quantity

🧠 Full Solution Path