A 16 pF capacitor is connected to 70 V supply. The amount of electric energy stored in the capacitor is

Options

A. `4.5xx10^(-12)J`
B. `5.1xx10^(-8)J`
C. `2.5xx10^(-12)J`
D. `3.92xx10^(-8)J`

Explanation

To find the amount of electric energy stored in a capacitor, we can use the formula:

\[ U = \frac{1}{2} C V^2 \]

where:
- \( U \) is the energy stored in the capacitor (in joules),
- \( C \) is the capacitance (in farads),
- \( V \) is the voltage (in volts).

**Step 1: Identify the values given in the problem.**
- The capacitance \( C = 16 \, \text{pF} = 16 \times 10^{-12} \, \text{F} \)
- The voltage \( V = 70 \, \text{V} \)

**Step 2: Substitute the values into the energy formula.**
\[ U = \frac{1}{2} \times (16 \times 10^{-12}) \times (70)^2 \]

**Step 3: Calculate \( V^2 \).**
\[ V^2 = 70^2 = 4900 \]

**Step 4: Substitute \( V^2 \) back into the energy formula.**
\[ U = \frac{1}{2} \times (16 \times 10^{-12}) \times 4900 \]

**Step 5: Calculate the product.**
\[ U = \frac{1}{2} \times 16 \times 4900 \times 10^{-12} \]
\[ = 8 \times 4900 \times 10^{-12} \]

**Step 6: Calculate \( 8 \times 4900 \).**
\[ 8 \times 4900 = 39200 \]

**Step 7: Final calculation.**
\[ U = 39200 \times 10^{-12} = 3.92 \times 10^{-8} \, \text{J} \]

Thus, the amount of electric energy stored in the capacitor is:

\[ \boxed{3.92 \times 10^{-8} \, \text{J}} \]

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