12th Grade > Physics
CAPACITANCE MCQs
Total Questions : 26
| Page 1 of 3 pages
Answer: Option C. -> 6
:
C
C=ϵ0Ad;C1=kϵ0A(d2)
C1C=2k;
12=2k;k=122=6
:
C
C=ϵ0Ad;C1=kϵ0A(d2)
C1C=2k;
12=2k;k=122=6
Answer: Option C. -> 6
:
C
Q1=Q2+Q3because in series combination charge is same on both the
condenser and V=V1+V2because in parallel combination V2=V3
HenceV=V1+V2
:
C
Q1=Q2+Q3because in series combination charge is same on both the
condenser and V=V1+V2because in parallel combination V2=V3
HenceV=V1+V2
Answer: Option B. -> 4 V
:
B
q=C1V1=C1V2
V1V2=C1C1=94.5=21
V1+V2V2=2+11
V2=123=4V
:
B
q=C1V1=C1V2
V1V2=C1C1=94.5=21
V1+V2V2=2+11
V2=123=4V
Answer: Option B. -> 66.7%
:
B
C1=ϵ0Ad;C2=ϵ0Ad−t+tk=ϵ0Ad2+d10
C2=106ϵ0Ad=5ϵ0A3d
C2C1=53;(C2C1−1)100%=(53−1)100%
=2×33.3%
=66.67%
:
B
C1=ϵ0Ad;C2=ϵ0Ad−t+tk=ϵ0Ad2+d10
C2=106ϵ0Ad=5ϵ0A3d
C2C1=53;(C2C1−1)100%=(53−1)100%
=2×33.3%
=66.67%
Answer: Option D. -> 1475 J
:
D
Energy = 12ϵ0E2×(A×d)=12ϵ0(V2d2)Ad
=12×8.85×10−12×(105)2×25×1060.75×103=1475J
:
D
Energy = 12ϵ0E2×(A×d)=12ϵ0(V2d2)Ad
=12×8.85×10−12×(105)2×25×1060.75×103=1475J
Answer: Option C. -> πC4
:
C
C=ϵ0l2d−−−−−−(1)
C1=ϵ0πl24d−−−−−−(2)
C1C=π4
C1=π4c
:
C
C=ϵ0l2d−−−−−−(1)
C1=ϵ0πl24d−−−−−−(2)
C1C=π4
C1=π4c
Answer: Option D. -> 550 μc
:
D
q1=c1V=5×10−6×10=50μc
q11=kc1v=12×5×10−6×10=600μc
Add.Charge=q11−q1=600−50
=550μC
:
D
q1=c1V=5×10−6×10=50μc
q11=kc1v=12×5×10−6×10=600μc
Add.Charge=q11−q1=600−50
=550μC
Answer: Option D. -> 4q3
:
D
Initial charge on sphere of radius R = q
Charge on this sphere after joining =q′=Rq+(−2q)×RR+2R=−q×R3R=−q3
So charge flown between them =q−(−q3)=4q3
:
D
Initial charge on sphere of radius R = q
Charge on this sphere after joining =q′=Rq+(−2q)×RR+2R=−q×R3R=−q3
So charge flown between them =q−(−q3)=4q3