b) A dielectric of dielectric constant 5.0 is filled in the gap between the plates of acapacitor. Calculate the factor by which the capacitance is increased, if thedielectric is only sufficient to fill up 1/5 of the gap. About the author Emma
Answer: C 0 = d ϵ 0 S C=C_1+C_2=\frac{\epsilon\epsilon_0(1/5)S}{d}+\frac{\epsilon_0(4/5)S}{d}=C=C 1 +C 2 = d ϵϵ 0 (1/5)S + d ϵ 0 (4/5)S = =\frac{\epsilon_0S}{d}(\epsilon/5+4/5)= d ϵ 0 S (ϵk=\frac{C}{C_0}=\epsilon/5+4/5=5/5+4/5=1.8k= C 0 C =ϵ/5+4/5=5/5+4/5=1.8 Explanation: PLEASE MARK ME AS BRAINLIST Reply
Answer:
C
0
=
d
ϵ
0
S
C=C_1+C_2=\frac{\epsilon\epsilon_0(1/5)S}{d}+\frac{\epsilon_0(4/5)S}{d}=C=C
1
+C
2
=
d
ϵϵ
0
(1/5)S
+
d
ϵ
0
(4/5)S
=
=\frac{\epsilon_0S}{d}(\epsilon/5+4/5)=
d
ϵ
0
S
(ϵk=\frac{C}{C_0}=\epsilon/5+4/5=5/5+4/5=1.8k=
C
0
C
=ϵ/5+4/5=5/5+4/5=1.8
Explanation:
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Answer:
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