Question

 A capacitor of capacitance (4.0 ± 0.2) μF is charged to a potential of (10.0 ± 0.1) V. The charge on the capacitor is:

• A. \( 2.5 \ \mu C \pm 3\% \)
• B. \( 2.5 \ \mu C \pm 6\% \)
• C. \( 40 \ \mu C \pm 3\% \)
• D. \( 40 \ \mu C \pm 6\% \)

Hint:

When calculating uncertainties for multiplication or division, sum up the relative percentage uncertainties of each term.

Answer 1

The Correct Option is D

Step 1: Using the Charge Formula 
The charge on a capacitor is given by the formula: \[ Q = C \times V. \] Substituting the given values: \[ Q = (4.0 \pm 0.2) \times (10.0 \pm 0.1) \ \mu C. \] Step 2: Calculating Charge 
\[ Q = 4.0 \times 10.0 = 40.0 \ \mu C. \] Step 3: Calculating Percentage Uncertainty 
The percentage uncertainties in \( C \) and \( V \) are: \[ \frac{\Delta C}{C} \times 100 = \frac{0.2}{4.0} \times 100 = 5\%, \] \[ \frac{\Delta V}{V} \times 100 = \frac{0.1}{10.0} \times 100 = 1\%. \] Since \( Q = C \times V \), the total percentage uncertainty is: \[ \text{Total Percentage Uncertainty} = 5\% + 1\% = 6\%. \] Step 4: Conclusion 
Thus, the charge on the capacitor is: \[ \mathbf{40 \ \mu C \pm 6\%}. \]