This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
The energy stored in a capacitor is given by the formula: E = (Q^2)/(2C) where E is the energy stored, Q is the charge on the plates, and C is the capacitance. Step 1: Identify the initial conditions. Initial charge Q_1 = 20 \, C Initial energy E_1 = 40 \, J Step 2: Determine the target energy. The question asks for the charge when the capacitor stores half the energy. Target energy E_2 = (1)/(2) E_1 = (1)/(2) (40 \, J) = 20 \, J. Step 3: Use the relationship between energy and charge. Since the capacitance C of the capacitor remains constant, the energy stored is directly proportional to the square of the charge (E Q^2). We can set up a ratio: (E_2)/(E_1) = (Q_2^2)/(Q_1^2) Step 4: Substitute the known values into the ratio. 20 \, J40 \, J = (Q_2^2)/((20 \, C))^2 (1)/(2) = (Q_2^2)/(400 \, C)^2 Step 5: Solve for Q_2^2. Q_2^2 = (1)/(2) × 400 \, C^2 Q_2^2 = 200 \, C^2 Step 6: Calculate Q_2. Q_2 = sqrt(200) \, C Q_2 ≈ 14.142 \, C Step 7: Compare with the given options. (a) 10.0C (b) 5.0C (c) 14.4C (d) 20.0C The calculated value 14.142 \, C is closest to 14.4 \, C. The final answer is 14.4C. That's 2 down. 3 left today — send the next one.