A parallel-plate capacitor of area 0.040 m^2 carries a charge q = 4.0 times 10^-

Question

A parallel-plate capacitor of area 0.040 m^2 carries a charge q = 4.0 times 10^-8 C. The potential across the plates increases with t according to the equation V = 50.0 mV + (0.10 mV/s)t, as a result of a time-dependent increase of die separation between the Plates. Find the function of time that describes the separation, A parallel-plate capacitor has square plates 6 cm on a side. Separated by 0.3 mm. The capacitor is charged to 3 V. then disconnected from the charging power supply. What is the charge density on the plates? The total charge on each plate? A capacitor consists of two parallel plates, each of area A. It is charged using a battery of potential V_0, which is then disconnected. (a) How much does the energy of the capacitor change it the separation of the plates is changed from d_0 to d_1? (b) How much work is done by the external force used to move the plates? (c) Suppose that the plates of the capacitor remain connected to the batters as they are moved. How much does the energy stored in the capacitor change under these conditions? (d) Is this change related to the work done by the force moving the plates? The electric field in a large thunderstorm is 125.000 V m. How much energy is contained in 1 m^3? In 1 km^3? The energy density in the space between the plates of a parallel-plate capacitor is 10^-6J/m^3. What is the voltage between the plates if the separation of the plates is 1 cm?

Explanation / Answer

we know that,

V = q/C

(50+0.1t)/1000= q/(Aeo/d) = 4e-8*d/(0.040*8.85e-12)

50 + 0.1t = 4e-5*d/(0.040*8.85e-12)

50 + 0.1t = 112994350 d

d = 50/112994350 + 0.1t/112994350

= ( 4.425*10^-7 + 8.85*10^-10 t)m/s

  

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