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 .

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

its a 5.5 n

glad that