TG EAMCET 3 May 2025 Shift 2 Paper

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=20µF
Step 2: Find Charge on the Capacitor
The capacitor is charged to 100 V . The charge stored is:
Q=Ci×V=20×10−6F×100V=2×10−3C
Step 3: Work Out Initial Energy With Dielectric
The energy stored to start with (with dielectric in place) is:
Ui=‌

Q2

2Ci

=‌

(2×10−3)2

2×20×10−6

=‌

4×10−6

40×10−6

=0.1J
Step 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=‌

Q2

2Cf

=‌

(2×10−3)2

2×4×10−6

=‌

4×10−6

8×10−6

=0.5J
Step 6: Work Done In Removing the Dielectric
Work done is the increase in energy:
W=Uf−Ui=0.5−0.1=0.4J