Can someone please help me solve PART D please! A dielectric-filled parallel-pla

Question

Can someone please help me solve PART D please!

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm^2, plate separation d = 5.00 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V. Throughout the problem, use epsilon_0 = 8.85 times 10^-12 C^2/N middot m^2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_3. U_3 = 1.66 times 10^-9 J In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric? W = 4.2 mitto 10^-10

Explanation / Answer

Work done = change in energy = Final energy - initial energy

= 1.66e-9 - 0.5*12.5^2*[3*8.85e-12*30e-4/5e-3]

= 4.15*10^-10 J ANSWER

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