The energy density u (defined as energy per unit volume) stored in an electrostatic field … — JAMB UTME

The energy density $u$ (defined as energy per unit volume) stored in an electrostatic field is related to the permittivity of free space $\epsilon_0$ and the electric field strength $E$ by the dimensional formula $u = k \epsilon_0^x E^y$ , where $k$ is a dimensionless constant. What is the value of the numerical exponent $y$ ?

Answer: 2

Answer

The value of the exponent

y

is 2.

By writing the dimensions of energy density

[M L^{-1} T^{-2}]

, permittivity

[M^{-1} L^{-3} T^4 I^2]

, and electric field strength

[M L T^{-3} I^{-1}]

, equating powers of electric current

I

yields

2x - y = 0

(or

y = 2x

). Substituting this into the equation for powers of mass

M

-x + y = 1

, yields

-x + 2x = 1

, so

x = 1

and

y = 2

Step-by-Step Solution

Derive the dimensional formulas for energy density

u

, permittivity

\epsilon_0

, and electric field

E

[u] = M L^{-1} T^{-2}, [\epsilon_0] = M^{-1} L^{-3} T^4 I^2, [E] = M L T^{-3} I^{-1}.

Expressing quantities in terms of base dimensions (M, L, T, I) is required for dimensional homogeneity.

Form the dimensional equation

u = k \epsilon_0^x E^y

and combine powers.

M L^{-1} T^{-2} = M^{-x+y} L^{-3x+y} T^{4x-3y} I^{2x-y}.

Applies the principle of dimensional consistency across the formula.

Equate corresponding powers of base dimensions to set up equations for

x

and

y

For I: 2x - y = 0; for M: -x + y = 1.

Base unit exponents on both sides of a physically valid equation must match.

Solve the algebraic equations for the unknown exponent

y

x = 1, y = 2.

Substituting y = 2x into -x + y = 1 directly gives x = 1 and y = 2.

Key Concept

Dimensional Analysis and Dimensional Homogeneity

Estimated Time:1m 30s

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