Question 10 of 11beginner🔧 ApplyNumerical2 marks

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Correct Answer

Given: Capacitance, C=12 pF=12×10−12 FC = 12 \text{ pF} = 12 \times 10^{-12} \text{ F} Potential difference, V=50 VV = 50 \text{ V}

Formula: The electrostatic energy (UU) stored in a capacitor is given by the formula:

U=12CV2U = \frac{1}{2}CV^2

Calculation: Substituting the given values into the formula:

U=12×(12×10−12 F)×(50 V)2U = \frac{1}{2} \times (12 \times 10^{-12} \text{ F}) \times (50 \text{ V})^2 U=12×12×10−12×2500U = \frac{1}{2} \times 12 \times 10^{-12} \times 2500 U=6×10−12×2500U = 6 \times 10^{-12} \times 2500 U=15000×10−12 JU = 15000 \times 10^{-12} \text{ J} U=1.5×10−8 JU = 1.5 \times 10^{-8} \text{ J}

Therefore, the electrostatic energy stored in the capacitor is 1.5×10−8 J1.5 \times 10^{-8} \text{ J}.

Exercise: EXERCISES | Q: 2.10 | (Page 80)
For More Understanding

Explanation

The solution follows the method shown in Example 2.10 of the provided text. The energy stored in a charged capacitor is defined by the relation U=12CV2U = \frac{1}{2}CV^2 (also listed as U=12QVU = \frac{1}{2}QV or Q22C\frac{Q^2}{2C} in the text). Since the capacitance and voltage are provided, the formula U=12CV2U = \frac{1}{2}CV^2 is the most direct application. The unit conversion from pF to F (1 pF=10−12 F1 \text{ pF} = 10^{-12} \text{ F}) is essential for standard unit consistency, as demonstrated in the textbook example.

Solution Steps

  1. Step 1: Convert capacitance to SI units: C=12 pF=12×10−12 FC = 12 \text{ pF} = 12 \times 10^{-12} \text{ F}.

  2. Step 2: Write the formula for energy stored: U=12CV2U = \frac{1}{2}CV^2.

  3. Step 3: Substitute the values C=12×10−12 FC = 12 \times 10^{-12} \text{ F} and V=50 VV = 50 \text{ V} into the formula.

  4. Step 4: Calculate the final value: U=1.5×10−8 JU = 1.5 \times 10^{-8} \text{ J}.

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