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When a potential difference of 150 V is applied to the plates of parallel-plate capacitor, the plates carry a surface charge density of 30 nc/cm^2. The spacing between the plates (in μm) is: a) 2.2 b) 1.1 c) 4.4 d) 6.6

Respuesta :

Karabo99

Answer:

s = 4.425×10^-6 μm

Explanation:

C = Q/V

but also:

C = ∈×A/s

for A being the area of the plates, the :

Q = 30 nc/(cm^-2)

then:

Q = (30×10^-9)×A/V = ∈×A/s

∈ = 8.85 pF/m

that is: (30×10^-9)/V = (8.85×10^-12)/s

then, s = (8.85×10^-12)(150)/(30×10^-9)

           = 4.425×10^-6 μm

Therefore, the spacing between the plates is 4.425×10^-6 μm.

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