A parallel-plate capacitor carries a constant charge Q (i.e., + Q on one plate,

Question

A parallel-plate capacitor carries a constant chargeQ (i.e., +Q on one plate, and -Q on theother). When a dielectric is inserted between the plates, whathappens to the voltage difference between the plates, and thepotential energy stored in the capacitor? The voltage and potential energy bothincrease.
None of thesechoices.
The voltageincreases, and the potential energy decreases.
The voltagedecreases, and the potential energy increases.
The voltageand potential energy both decrease.
A parallel-plate capacitor is attached to abattery that maintains a constant voltage differenceV across the plates. When a dielectric is insertedbetween the plates, what happens to the charge on the plates, andthe potential energy stored in the capacitor? The charge increases, and the potential energydecreases.
The chargeand potential energy both decrease.
None of thesechoices.
The chargedecreases, and the potential energy increases.
The chargeand potential energy both increase. The voltage and potential energy bothincrease.
None of thesechoices.
The voltageincreases, and the potential energy decreases.
The voltagedecreases, and the potential energy increases.
The voltageand potential energy both decrease.
A parallel-plate capacitor is attached to abattery that maintains a constant voltage differenceV across the plates. When a dielectric is insertedbetween the plates, what happens to the charge on the plates, andthe potential energy stored in the capacitor? The charge increases, and the potential energydecreases.
The chargeand potential energy both decrease.
None of thesechoices.
The chargedecreases, and the potential energy increases.
The chargeand potential energy both increase.

Explanation / Answer

   Q1)             Ans:The voltage and potential energy both decrease.                        Reason: When the dielectric is inserted, then thevoltage decrease by k timess andcapacity of                           thecapacitor increase by k times                          Energy , E = (1/2 ) C V2                          Hence, energy decrease to k times theoriginal    Q2)             Ans:The charge and potential energy both increase.                        Reason: When the dielectric is inserted, then the capacityof  the capacitor and also the             chargeincrease by k times                          Energy , E = (1/2 ) ( Q2/ C  )                        Hence, energy increase to k times theoriginal                          Hence, energy decrease to k times theoriginal    Q2)             Ans:The charge and potential energy both increase.                        Reason: When the dielectric is inserted, then the capacityof  the capacitor and also the             chargeincrease by k times                          Energy , E = (1/2 ) ( Q2/ C  )                        Hence, energy increase to k times theoriginal                        Reason: When the dielectric is inserted, then the capacityof  the capacitor and also the             chargeincrease by k times                          Energy , E = (1/2 ) ( Q2/ C  )                        Hence, energy increase to k times theoriginal                          Energy , E = (1/2 ) ( Q2/ C  )                        Hence, energy increase to k times theoriginal

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