An initially uncharged air-filled capacitor is connected to a 2.67 V2.67 V charging source. As a result, the capacitor acquires 6.39×10−5 C6.39×10−5 C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant κ of this substance is 6.03.6.03. Find the voltage VV across the capacitor and the charge fQf stored by it after the dielectric is inserted and the circuit has returned to a steady state.

Explanation:

capacitance of air  capacitor C = charge / volt

C = 6.39 x 10⁻⁵ / 2.67

= 2.39 x 10⁻⁵ Farad

After inserting dielectric between plates , the capacitance of capacitor becomes C₁ where

C₁ = 6.03 x 2.39 x 10⁻⁵

= 14.41 x 10⁻⁵ Farad

The voltage of the capacitor will remain the same as earlier that is 2.67 V because it remains connected with the charging source .

New charge on the capacitor = new capacitance x voltage

= 14.41 x 10⁻⁵ x 2.67

= 38.47 x 10⁻⁵ C .

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

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