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प्रश्न
Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.
- Calculate the capacitance and the rate of charge of the potential difference between the plates.
- Obtain the displacement current across the plates.
- Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
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उत्तर
Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε0 = 8.854 × 10−12 C2 N−1 m−2
(a) Capacitance between the two plates is given by the relation,
C = `(ε_0"A")/"d"` .....[∵ A = πr2 = 3.14 × (0.12)2]
= `(8.854 xx 10^-12 xx 3.14 xx (0.12)^2)/0.05`
= 8.01 × 10−12 F
= 8.01 pF
Charge on each plate, q = CV ⇒ V = `"q"/"C"`
`therefore "dV"/"dt" = 1/"C" "dq"/"dt" = 1/"C" "I" ...(because "dq"/"dt" = "I")`
Rate of charge of the potential difference `"dV"/"dt"` = `0.15/(8.01 xx 10^-12)`
= 1.875 × 109 V s−1
(b) Displacement current across the plates,
`"I"_"d" = epsilon_0 ("d"phi_"E")/"dt"`
Where `phi_"E"` is the electric flux passing through a closed loop between the plates.
∵ The electric field E between the plates = `"q"/(epsilon_0 "A")`
∴ If the area of the loop is A then,
`phi_"E" = oint vec("E") * "d" vec("A") = oint "E dA" ....[because vec("E") ⊥ "d" vec("A")]`
`=> phi_"E" = "EA" = "q"/epsilon_0 => ("d"phi_"E")/"dt" = 1/epsilon_0 * "dq"/"dt"`
∴ `"I"_"d" = epsilon_0 1/epsilon_0 "dq"/"dt" = "I"`
⇒ Displacement current, `"I"_"d"` = 0.15 A
(c) Yes, Kirchhoff's first law is very much applicable to each plate of capacitor as Id = I.
So current is continuous and constant across each plate.
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