Step 1: Find Initial Capacitance
The original capacitor has air and its capacitance is
4μF. When a dielectric with constant 5 fills the gap, the new capacitance becomes 5 times bigger:
Ci​=5×4μF=20μFStep 2: Find Charge on the Capacitor
The capacitor is charged to 100 V . The charge stored is:
Q=Ci​×V=20×10−6F×100V=2×10−3CStep 3: Work Out Initial Energy With Dielectric
The energy stored to start with (with dielectric in place) is:
Ui​=2Ci​Q2​=2×20×10−6(2×10−3)2​=40×10−64×10−6​=0.1JStep 4: Find Final Capacitance After Removing Dielectric
When the dielectric is taken out, the capacitance returns to air value:
Cf​=4μF.
Step 5: Calculate Final Energy With Dielectric Gone
The charge on the plates stays the same because the capacitor is disconnected from the battery. The energy now is:
Uf​=2Cf​Q2​=2×4×10−6(2×10−3)2​=8×10−64×10−6​=0.5JStep 6: Work Done In Removing the Dielectric
Work done is the increase in energy:
W=Uf​−Ui​=0.5−0.1=0.4J