A capacitor of capacitance `C = 2.0 +- 0.1 mu F` is charged to a voltage `V = 20+- 0.2 V`. What will be the charge `Q` on the capacitor ? Use `Q = CV`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the charge \( Q \) using the formula \( Q = CV \).

Given:
- Capacitance \( C = 2.0 \, \mu F = 2.0 \times 10^{-6} \, F \)
- Voltage \( V = 20 \, V \)

Using the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q = (2.0 \times 10^{-6} \, F) \times (20 \, V) = 40 \times 10^{-6} \, C = 40 \, \mu C \]

### Step 2: Calculate the uncertainty in charge \( \Delta Q \).

To find the uncertainty in the charge, we use the formula for the propagation of uncertainties:
\[ \frac{\Delta Q}{Q} = \frac{\Delta C}{C} + \frac{\Delta V}{V} \]

Where:
- \( \Delta C = 0.1 \, \mu F = 0.1 \times 10^{-6} \, F \)
- \( \Delta V = 0.2 \, V \)

Now substituting the values:
\[ \frac{\Delta C}{C} = \frac{0.1 \times 10^{-6}}{2.0 \times 10^{-6}} = \frac{0.1}{2.0} = 0.05 \]
\[ \frac{\Delta V}{V} = \frac{0.2}{20} = 0.01 \]

Now, adding these two uncertainties:
\[ \frac{\Delta Q}{Q} = 0.05 + 0.01 = 0.06 \]

### Step 3: Calculate \( \Delta Q \).

Now, we can find \( \Delta Q \):
\[ \Delta Q = Q \times \frac{\Delta Q}{Q} = 40 \, \mu C \times 0.06 = 2.4 \, \mu C \]

### Step 4: Write the final result for the charge \( Q \) with uncertainty.

Thus, the charge on the capacitor is:
\[ Q = 40 \, \mu C \pm 2.4 \, \mu C \]

### Final Answer:
The charge on the capacitor is \( 40 \, \mu C \pm 2.4 \, \mu C \).

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