Question

\(\frac{\text{Force} \times \text{Length}}{\text{Charge}}=\)which quantity ?

• A. Potential
• B. Current
• C. Resistance
• D. Capacity

Answer 1

The Correct Option is A

To determine the quantity represented by $\frac{\text{Force} \times \text{Length}}{\text{Charge}}$, let's analyze the units and physical meaning step by step.

Step 1: Identify the Units of Each Quantity
1. Force: The unit of force is Newton ($\text{N}$).
- In terms of base SI units: $1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$.
2. Length: The unit of length is meter ($\text{m}$).
3. Charge: The unit of charge is Coulomb ($\text{C}$).

Step 2: Write the Expression in Terms of Units
The given expression is:
$$ \frac{\text{Force} \times \text{Length}}{\text{Charge}} $$
Substituting the units:
$$ \frac{\text{N} \cdot \text{m}}{\text{C}} = \frac{(\text{kg} \cdot \text{m/s}^2) \cdot \text{m}}{\text{C}} = \frac{\text{kg} \cdot \text{m}^2/\text{s}^2}{\text{C}} $$

Step 3: Interpret the Physical Meaning
The term $\frac{\text{Force} \times \text{Length}}{\text{Charge}}$ represents the work done per unit charge, which is the definition of electric potential (or voltage). Electric potential is measured in volts ($\text{V}$), where:
$$ 1 \, \text{V} = 1 \, \text{J/C} = 1 \, \text{kg} \cdot \text{m}^2/(\text{s}^2 \cdot \text{C}) $$

Step 4: Match with the Options
From the options provided:
- (1) Potential: Correct, as it matches the derived unit.
- (2) Current: Incorrect, as current has units of Amperes ($\text{A}$).
- (3) Resistance: Incorrect, as resistance has units of Ohms ($\Omega$).
- (4) Capacity: Incorrect, as capacity typically refers to capacitance, which has units of Farads ($\text{F}$).

Final Answer: $ {\text{Potential}} $