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Difficulty: HardFundamental and Derived Units

The derived SI unit of electric potential, the volt (V\text{V}), can be expressed in terms of fundamental SI base units as kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d. What is the value of the exponent cc?

Answer: -3

Answer

The exponent cc corresponding to the base unit second (s\text{s}) is −3-3.
Electric potential difference (VV) is defined as work (WW) per unit charge (QQ). In terms of SI base units, work has units of kg⋅m2⋅s−2\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} and charge has units of A⋅s\text{A} \cdot \text{s}. Dividing work by charge gives kg⋅m2⋅s−2A⋅s=kg1⋅m2⋅s−3⋅A−1\frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}} = \text{kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}. Equating this to kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d yields c=−3c = -3.

Step-by-Step Solution

1
Relate electric potential to work and charge
1 V=1 J1 C\text{1 V} = \frac{\text{1 J}}{\text{1 C}}
Electric potential difference is defined as work done per unit electric charge.
2
Break down Joules into base SI units
1 J=1 N⋅m=(1 kg⋅m⋅s−2)⋅m=1 kg⋅m2⋅s−2\text{1 J} = \text{1 N} \cdot \text{m} = (\text{1 kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m} = \text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}
Work is force multiplied by displacement, and force is mass multiplied by acceleration.
3
Break down Coulombs into base SI units
1 C=1 A⋅s\text{1 C} = \text{1 A} \cdot \text{s}
Electric charge is electric current multiplied by time.
4
Substitute base unit expressions into the ratio for electric potential
1 V=1 kg⋅m2⋅s−21 A⋅s=1 kg1⋅m2⋅s−3⋅A−1\text{1 V} = \frac{\text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{1 A} \cdot \text{s}} = \text{1 kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}
Combining powers of base units yields the complete fundamental representation.
5
Determine the exponent value cc
c=−3c = -3
Comparing kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d to kg1⋅m2⋅s−3⋅A−1\text{kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} shows that the exponent of s\text{s} is −3-3.

Key Concept

Expressing derived units in terms of fundamental SI base units
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Topics

Fundamental and Derived Units

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