A parallel-plate capacitor is charged by a battery of voltage V0. The battery is then completely disconnected, and a dielectric slab of dielectric constant K is slowly inserted to completely fill the space between the plates. Which of the following statements correctly describes the changes in the system?
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Correct Answer
Charge remains constant, capacitance becomes K times, voltage decreases by K times, electric field decreases by K times, and stored energy decreases by K times.
Step-by-Step Explanation
The most important point is that the battery is disconnected before inserting the dielectric.
Therefore, the capacitor is isolated and no charge can flow onto or away from the plates.
Q = constant
Initially, without dielectric:
C0 = ε0A/d
Since the dielectric completely fills the space, the new capacitance becomes:
C = K C0
Because Q remains constant:
V = Q/C
Therefore:
V = V0/K
What Happens to Electric Field?
For a parallel-plate capacitor:
E = V/d
Since the plate separation d is unchanged and voltage becomes V0/K:
E = E0/K
Thus, the electric field decreases by a factor of K.
What Happens to Stored Energy?
For an isolated capacitor, charge is constant. Hence we use:
U = Q²/(2C)
Since C becomes KC0:
U = U0/K
So the stored electrostatic energy decreases by a factor of K.
The decrease in electrostatic energy is associated with the work done by the electric field while the dielectric is pulled into the capacitor.
Final Changes at a Glance
| Quantity | Initially | After dielectric | Change |
|---|---|---|---|
| Charge Q | Q0 | Q0 | Constant |
| Capacitance C | C0 | KC0 | Increases K times |
| Voltage V | V0 | V0/K | Decreases K times |
| Electric field E | E0 | E0/K | Decreases K times |
| Energy U | U0 | U0/K | Decreases K times |
JEE Quick Trick
Battery disconnected → Q constant.
Then remember:
C ↑ K times → V ↓ K times → E ↓ K times → U ↓ K times.
Then remember:
C ↑ K times → V ↓ K times → E ↓ K times → U ↓ K times.
30 Related MCQs with Solutions
1. A charged capacitor is disconnected from the battery. A dielectric of constant K is inserted completely. The charge becomes:
A. KQ
B. Q/K
C. Q
D. Zero
Answer: C. Q
2. The capacitance after completely inserting dielectric is:
A. C/K
B. C
C. KC
D. K²C
Answer: C. KC
3. The voltage across the isolated capacitor becomes:
A. KV0
B. V0/K
C. V0
D. K²V0
Answer: B. V0/K
4. The electric field between the plates becomes:
A. KE0
B. E0/K
C. E0
D. K²E0
Answer: B. E0/K
5. The energy stored in the isolated capacitor becomes:
A. KU0
B. U0/K
C. U0
D. K²U0
Answer: B. U0/K
6. Which quantity remains constant after battery disconnection?
A. Voltage
B. Capacitance
C. Charge
D. Electric field
Answer: C. Charge
7. If K = 4, the final capacitance is:
A. C/4
B. 2C
C. 4C
D. C
Answer: C. 4C
8. If K = 5, the final voltage is:
A. 5V0
B. V0/5
C. V0
D. 25V0
Answer: B. V0/5
9. If K = 3, the electric field becomes:
A. 3E0
B. E0/3
C. E0
D. 9E0
Answer: B. E0/3
10. For an isolated capacitor, the appropriate energy formula is:
A. U = ½CV²
B. U = Q²/(2C)
C. U = CV
D. U = Q/C
Answer: B. U = Q²/(2C)
11. Why does charge remain constant?
A. Battery continues charging
B. Plates are isolated
C. Dielectric creates charge
D. Voltage source increases
Answer: B. Plates are isolated
12. If capacitance increases while Q is constant, voltage:
A. Increases
B. Decreases
C. Remains constant
D. Becomes infinite
Answer: B. Decreases
13. Dielectric insertion changes capacitance because:
A. Permittivity increases
B. Plate area becomes zero
C. Charge disappears
D. Plate separation doubles
Answer: A
14. The permittivity of the dielectric is:
A. ε0/K
B. Kε0
C. K²ε0
D. ε0
Answer: B. Kε0
15. For K > 1, dielectric insertion in an isolated capacitor causes capacitance to:
A. Increase
B. Decrease
C. Remain unchanged
D. Become zero
Answer: A
16. The relation Q = CV gives, for constant Q:
A. V ∝ C
B. V ∝ 1/C
C. V ∝ C²
D. V is independent of C
Answer: B. V ∝ 1/C
17. If K = 2, energy stored becomes:
A. 2U0
B. U0/2
C. 4U0
D. U0
Answer: B. U0/2
18. The force between plates of an isolated capacitor after dielectric insertion generally:
A. Increases
B. Decreases
C. Remains exactly unchanged
D. Becomes infinite
Answer: B. Decreases
19. Which condition is essential for Q to remain constant?
A. Battery remains connected
B. Battery is disconnected
C. Voltage is increased
D. Plate area is changed
Answer: B
20. If the battery were still connected, the voltage would:
A. Remain V0
B. Become V0/K
C. Become zero
D. Become KV0
Answer: A. V0
21. If the battery remains connected, charge on the capacitor becomes:
A. Q/K
B. KQ
C. Q
D. Zero
Answer: B. KQ
22. The key difference between connected and disconnected cases is:
A. Connected → V constant; disconnected → Q constant
B. Both have Q constant
C. Both have V constant
D. Neither changes
Answer: A
23. When dielectric completely fills the capacitor, capacitance is:
A. ε0A/d
B. Kε0A/d
C. ε0d/A
D. Kd/A
Answer: B
24. The electric field in a completely filled isolated capacitor is:
A. E0
B. KE0
C. E0/K
D. K²E0
Answer: C
25. The electrostatic energy density inside a dielectric-filled capacitor is:
A. Always increased by K
B. Reduced in this isolated case
C. Zero
D. Infinite
Answer: B
26. For the isolated capacitor, inserting dielectric causes V to change because:
A. Q changes
B. C changes while Q remains constant
C. Plate area becomes zero
D. Battery supplies more charge
Answer: B
27. Which relation is always valid for a capacitor?
A. Q = CV
B. Q = C/V
C. V = QC
D. C = QV
Answer: A. Q = CV
28. The dielectric reduces the electric field for an isolated capacitor because:
A. Polarization opposes the original field
B. Charge becomes infinite
C. Battery increases voltage
D. Plate separation becomes zero
Answer: A
29. Which statement is correct?
A. Disconnected capacitor → V constant
B. Disconnected capacitor → Q constant
C. Disconnected capacitor → C constant
D. Disconnected capacitor → U constant
Answer: B
30. If the dielectric constant is K, the final energy compared with initial energy is:
A. KU0
B. U0/K
C. K²U0
D. U0
Answer: B. U0/K
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