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NEET
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Physics
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Capacitance

NEET · Physics · Capacitance

A parallel plate capacitor is being charged such that a constant...

A parallel plate capacitor is being charged such that a constant displacement current of 8.85 mA8.85 \text{ mA} exists between its plates. If the area of each plate is 100 cm2100 \text{ cm}^2, what is the rate at which the electric field between the plates is changing?

(Given ε0=8.85×10−12 C2 N−1 m−2\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2})

Options

  1. A1015 V m−1 s−110^{15} \text{ V m}^{-1} \text{ s}^{-1}
  2. B107 V m−1 s−110^{7} \text{ V m}^{-1} \text{ s}^{-1}
  3. C8.85×1011 V m−1 s−18.85 \times 10^{11} \text{ V m}^{-1} \text{ s}^{-1}
  4. D1011 V m−1 s−110^{11} \text{ V m}^{-1} \text{ s}^{-1}

Correct answer

D. 1011 V m−1 s−110^{11} \text{ V m}^{-1} \text{ s}^{-1}

Step-by-step solution

The displacement current IdI_d is given by the fundamental relation:
Id=ε0dΦEdtI_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}

Since the electric flux ΦE=E×A\Phi_E = E \times A, we can write:
Id=ε0AdEdtI_d = \varepsilon_0 A \dfrac{dE}{dt}

Rearranging for the rate of change of electric field dEdt\dfrac{dE}{dt}:
dEdt=Idε0A\dfrac{dE}{dt} = \dfrac{I_d}{\varepsilon_0 A}

Given values:
Id=8.85 mA=8.85×10−3 AI_d = 8.85 \text{ mA} = 8.85 \times 10^{-3} \text{ A}
A=100 cm2=100×10−4 m2=10−2 m2A = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 = 10^{-2} \text{ m}^2
ε0=8.85×10−12 C2 N−1 m−2\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}

Substituting the values:
dEdt=8.85×10−38.85×10−12×10−2\dfrac{dE}{dt} = \dfrac{8.85 \times 10^{-3}}{8.85 \times 10^{-12} \times 10^{-2}}

dEdt=10−310−14=1011 V m−1 s−1\dfrac{dE}{dt} = \dfrac{10^{-3}}{10^{-14}} = 10^{11} \text{ V m}^{-1} \text{ s}^{-1}

Answer: 1011 V m−1 s−110^{11} \text{ V m}^{-1} \text{ s}^{-1}
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