capacitance of capacitor for conducting slab
Answers
Answer:
If the space between the plates is vacuum, the capacitance is given by \bold{\mathrm{C}_{2}=\epsilon_{0} \mathrm{A} / \mathrm{d}}C
2
=ϵ
0
A/d where d = plate separation, A = area
Now let’s consider that charges on the capacitor plates are ± Q. Then the uniform electric field set up between the capacitor plates is
E_{0}=\frac{\sigma}{c_{0}}=\frac{Q}{A \epsilon_{0}}E
0
=
c
0
σ
=
Aϵ
0
Q
where Ϭ is the surface charge density. The potential difference between the capacitor plates will be \mathrm{V}_{0}=\mathrm{E}_{0} \mathrm{d}=\mathrm{Qd} / \mathrm{A} \epsilon_{0}V
0
=E
0
d=Qd/Aϵ
0
When a conducting slab of thickness 1 < d is placed between the capacitor plates, free electrons flow inside it so as to reduce the field to zero inside the slab. Charges -Q and +Q appear on the upper and lower faces of the slab. Now electric field exists only in the vacuum regions between the plates of the capacitor on the either side of the slab i.e. the field exists only in thickness d – t, therefore, potential difference between the plates of the capacitor is
\bold{\mathrm{V}=\mathrm{E}_{0}(\mathrm{d}-1)=\mathrm{Q}(\mathrm{d}-1) / \mathrm{A} \epsilon_{0}}V=E
0
(d−1)=Q(d−1)/Aϵ
0
Hence capacitance of the capacitor in the presence of conducting slab becomes
\mathrm{C}=\mathrm{Q} / \mathrm{V}=\mathrm{A} \epsilon_{0} /(\mathrm{d}-1)=\frac{\mathrm{A} \cdot 0}{d} \times \frac{d}{d-1}=\frac{d}{d-1} \times \mathrm{Co}C=Q/V=Aϵ
0
/(d−1)=
d
A⋅0
×
d−1
d
=
d−1
d
×Co
Explanation:
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