A capacitor is connected up to a 12.0 V battery and allowed to charge up. Once t

Question

A capacitor is connected up to a 12.0 V battery and allowed to charge up. Once the capacitor is fully charged, the battery is removed and a material with a dielectric constant is removed from the region between the two plates. From the quantities listed below, select all those items that decrease. The capacitance of the capacitor The potential difference between the plates The charge on the plates The electric potential energy stored by the capacitor The electric field between the plates None of the above

Explanation / Answer

1) Capacitance of capacitor is dircetly proportional to dielectric constant. As dielectric is removed, Capacitance decrease by factor of k.
2) Potential difference between plates increase.
3) Charge on plates remains constant.
2 and 3 can be explained on the basis of fact that q = CV. As Baterry is removed from, there is no change in charge. With decrease in C ther is increase in V, potentila difference.

4) Energy stored in capacitor = q2 / 2C
As q is constant and C decrease, there is increase in energy stored.

5) Electric field E = V/d .
As there is increase in V , E increases.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.