A parallel plate capacitor of capacity 5uF and
plate separation 6cm is connected to a 1V
battery and is charged. A dielectric of
dielectric constant 4 and thickness 4 cm is
introduced into the capacitor. The additional
charge that flows into the capacitor from the
battery is
1) 2μα 2) 3°C 3) 5μα 4) 10μο
Answers
Answered by
2
Explanation:
Correct option is
C
5 μC
Charge on capacitor plates without the dielectric is
=CV=(5×10
−6
F)×1V=5×10
−6
C=5μC
The capacitance after the dielectric is introduced is
C
′
=
d−(t−
K
t
)
ε
0
A
=
1−
⎝
⎜
⎜
⎜
⎛
d
t−
K
t
⎠
⎟
⎟
⎟
⎞
ε
0
A/d
=
1−
⎝
⎜
⎜
⎜
⎛
d
t−
K
t
⎠
⎟
⎟
⎟
⎞
C
=
1−
⎝
⎜
⎜
⎜
⎛
6cm
4cm−
4
4cm
⎠
⎟
⎟
⎟
⎞
5μF
=
1−(
6
4−1
)
5μF
=10μF
∴ Charge on capacitor plate now will be
Q
′
=C
′
V=10μF×1V=10μC
Additional charge transferred =Q
′
−Q=10μC−5μC=5μC
Answered by
0
Answer:
5μα is correct answer
pl mark has brainly
Similar questions