This set of Class 12 Physics Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Effect of Dielectric on Capacitance”.
1. What is the dielectric constant of the medium if the capacitance of a parallel plate capacitor increases from 40F to 80F on introducing a dielectric medium between the plates?
a) 20
b) 0.5
c) 2
d) 5
View Answer
Explanation: Capacitance without dielectric = 40 F.
Capacitance with dielectric = 80 F.
k = \(\frac {80}{40}\)
k = 2.
2. How does the potential difference change with the effect of the dielectric when the battery is kept disconnected from the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
View Answer
Explanation: When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field. The reduction in the electric field results in a decrease in potential difference.
V = Ed = \(\frac {E_0 d}{k}=\frac {V_0}{k}\).
3. How does the potential difference change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
View Answer
Explanation: As the battery remains connected across the capacitor, so the potential difference remains constant at V0 even after the introduction of the dielectric slab. In this way, dielectric has an effect on potential difference.
4. How does the capacitance change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Zero
d) Remains constant
View Answer
Explanation: When a dielectric is introduced, and the battery remains connected across the capacitor, the capacitance increases from C0 to C.
C = kC0.
5. How does the electric field change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Remains unchanged
d) Zero
View Answer
Explanation: As the potential difference remains unchanged, so the electric field E0 between the capacitor plates remain unchanged.
E = \(\frac {V}{d}\) = \(\frac {V_0}{d}\) = E0.
6. The charge on the capacitor plates decreases from Q0 to Q with the effect of the dielectric when the battery remains connected across the capacitor.
a) True
b) False
View Answer
Explanation: When a dielectric is introduced, and he battery remains connected across the capacitor, the charge on the capacitor plates increases from Q0 to Q.
Q = CV = kC0.V0 = kQ0.
7. How does the capacitance change with the effect of the dielectric when the battery is kept disconnected from the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Zero
View Answer
Explanation: When the battery is disconnected, and a dielectric is introduced, there will be a decrease in potential difference and as a result, the capacitance increases k times.
C = \(\frac {Q_0}{V}\)
C = \( [ \frac {Q_0}{(\frac {V_0}{k})} ] \)
C = \(\frac {kQ_0}{V_0}\)
C = kC0.
8. In a parallel plate capacitor, the capacitance increases from 100 F to 800 F, on introducing a dielectric medium between the plates. What is the dielectric constant of the medium?
a) 0.125
b) 125
c) 80
d) 8
View Answer
Explanation: Capacitance with dielectric = 800 F
Capacitance without dielectric = 100 F
Dielectric constant = \( ( \frac {Capacitance \, with \, dielectric}{Capacitance \, without \, dielectric} ) \)
k = \(( \frac {800 F}{100 F} )\)
k = 8.
Therefore, the dielectric constant is calculated as 8.
Sanfoundry Global Education & Learning Series – Physics – Class 12.
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