A parallel plate capacitor of capacitance 8.0 A parallel plate capacitor of capa

Question

A parallel plate capacitor of capacitance 8.0

A parallel plate capacitor of capacitance 8.0 mu F has the space between the plates filled with a slab of glass with? = 2.2. The capacitor is charged by attaching it to a 1.2-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. Find the capacitance after the glass is removed. Find the potential difference after the glass is removed. Find the charge on the plates after the glass is removed. Find the energy stored in the capacitor after the glass is removed.

Explanation / Answer

The capacitance of a parallel-plate capacitor:

C = A * epsilon / d

where A is plate area and d is distance. epsilon is k * epsilon0.

In any capacitor:

Q = C * V

where C is capacitance, Q is stored charge, and V is voltage across the plates.

A: If k drops from 2.2 to 1, then C must drop from 8.0 to 3.636 uF
B: Use the second formula to find V, as Q remained constant. (I get 2.64 V)
C: I used the value of Q to find the above. (I got 9.599 ucoul.)
D: U = 1/2 C*V^2 = 1/2 Q*V. (I got 12.67 uJ)

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