The dimensional formula of ( = permittivity of vacuum and = electric field) is . The value of = ______. [2026]

Question

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 $2 a - b + c$ = ______. [2026]

Smart Achievers · Accepted Answer

(2)

$\frac{1}{2} ε_{0} E^{2} = \frac{d u}{d v} = \frac{Energy}{Volume}$

$∴$ $2 a - b + c$

$∴$ $a = 1, b = - 1, c = - 2$

$∴$ $2 a - b + c = 2 + 1 - 2 = 1$

Solution

Q.
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 $2 a - b + c$ = ______. [2026]

1 0

2 1

3 -1

4 2

Ans.
(2)

$\frac{1}{2} ε_{0} E^{2} = \frac{d u}{d v} = \frac{Energy}{Volume}$

$∴$ $[\frac{1}{2} ε_{0} E^{2}] = \frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}$

$∴$ $a = 1, b = - 1, c = - 2$

$∴$ $2 a - b + c = 2 + 1 - 2 = 1$

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Solution

Q. The dimensional formula of 12ε0E2 (ε0 = permittivity of vacuum and E = electric field) is MaLbTc. The value of 2a-b+c = ______. [2026]

1 0

2 1

3 -1

4 2

Ans. (2) 12ε0E2=dudv=EnergyVolume ∴ [12ε0E2]=ML2T-2L3=ML-1T-2 ∴ a=1, b=-1, c=-2 ∴ 2a-b+c=2+1-2=1

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Q.
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 $2 a - b + c$ = ______. [2026]

Ans.
(2)

$\frac{1}{2} ε_{0} E^{2} = \frac{d u}{d v} = \frac{Energy}{Volume}$

$∴$ $[\frac{1}{2} ε_{0} E^{2}] = \frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}$

$∴$ $a = 1, b = - 1, c = - 2$

$∴$ $2 a - b + c = 2 + 1 - 2 = 1$