Help me answer part D: Find the energy stored in the capacitor after the glass i

Question

Help me answer part D: Find the energy stored in the capacitor after the glass is removed. I used formula U=Q^2/2C0 and got 33.13 but that is incorrect. Help me find the answer in Microjoules please (uJ).

Round your answer(s) to three significant fugures. A parallel plate capacitor of capacitance 5.88 F has the space between the plates filled with a slab of glass withk -3.14. The capacitor is charged by attaching it to a 6.0-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed Part 1 (e) Find the capacitance after the glass is removerd. 1.87 Part 2 (b) Find the potential difference after the glass is removed. 18.8 Part 3 (c) Find the charge on the plates after the glass is removed. 35.2 HC

Explanation / Answer

You have to use the values of capacitance and voltage after the glass is removed to find energy stored.

Energy stored = 1/2*C*V2

Energy stored = 0.5*1.87*18.8*18.8

Energy stored. = 332.8 microjoules.

If I use the formula which you have used I get

Energy stored = 35.28*35.28/2*1.87

Energy stored = 332.8 microjoules.

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