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A parallel plate capacitor is charged by a battery, then the battery is disconnected. A dielectric slab with relative permittivity epsilon_r = 2 of the same area as the plates but with thickness equal to half the plate separation is then inserted between the plates. What is the percentage change in the stored potential energy?
- Decrease by 25%
- Decrease by 50%
- Increase by 25%
- Increase by 50%
Correct answer: Decrease by 25%
Solution
Let original capacitance C0 = epsilon₀ * A / d. After inserting slab of thickness d/2 with epsilon_r = 2: system is two capacitors in series. C1 = 2*epsilon₀*A/(d/2) = 4*epsilon₀*A/d = 4*C0. C2 = epsilon₀*A/(d/2) = 2*C0. New C = C1*C2/(C1+C2) = 4C0*2C0/(4C0+2C0) = 8C0²/(6C0) = 4C0/3. Energy before: U0 = Q²/(2*C0). Energy after: U = Q²/(2*(4C0/3)) = 3Q²/(8C0). Change: U - U0 = 3Q²/(8C0) - Q²/(2C0) = 3Q²/(8C0) - 4Q²/(8C0) = -Q²/(8C0). Percentage change = (-Q²/(8C0))/(Q²/(2C0)) * 100 = (-1/8)/(1/2) * 100 = -25%. Decrease by 25%.
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