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tfguss

  • 3 years ago

Please help! This homework is due tonight for me + I don't get it. A capacitor with air between its plates is charged to 150 V and then disconnected from the battery. When a piece of glass is placed between the plates, the voltage across the capacitor drops to 20 V. What is the dielectric constant of the glass? (Assume the glass completely fills the space between the plates.)

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  1. Jemurray3
    • 3 years ago
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    The voltage across the capacitor is V = Q/C where Q is the charge on the plates and C is the capacitance. If the battery is disconnected, the charge Q on the plates must be constant. The slab of glass changes the capacitance from C to kC where k is the dielectric constant of the glass. Therefore, before the glass is inserted, the voltage is V = Q/C. Afterwards, the new voltage V' = Q/(kC). From this, figure out what k must be.

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