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If you double the area of a parallel-plate capacitor and reduce the distance between the plates...
A parallel plate capacitor has a capacitance of 400uF. If the distance between the parallel plates is 15mm and the dielectric between the plates is air (k=1), what is the area of each parallel plate.
The Plates of a parallel plate capacitor are seperated by 1.0mm. (a)What is the area of the plate if the capacitance is 2.5(micro Farad) (b) if the distance of seperation is doubled and the plate size is halved, by what factor does the capacitance change? please explain your answer, thank you
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has charge of magnitude Qo on its plates. It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them. After the plates are 2d apart, by what factor does the capacitance change? By what factor does the potential difference between the plates change?
If you reduce the gap between a parallel-plate capacitor to } of the original distance while maintaining the same charge on plates, what is the ratio of the electric potential energy stored in the capacitor with the narrower gap over the energy stored in the wider gap?
Find the capacitance of a parallel plate capacitor having plates of area 5.75 m2 that are separated by 0.495 mm of Teflon. The dielectric constant of Teflon is 2.1 X How is capacitance related to plate area and separation for a parallel plate capacitor? Did you consider the effect of the dielectric? HF
A parallel plate capacitor has plates of area A = 5.50 ✕ 10−2 m2 separated by distance d = 1.32 ✕ 10−4 m. (The permittivity of free space is ε0 = 8.85 ✕ 10−12 C2/(N · m2).) (a) Calculate the capacitance (in F) if the space between the plates is filled with air. . What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant κ = 3.10 as in...
A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery and the space between the plates is filled halfway with a material having the dielectric constant K. What are the capacitance and the magnitude of the charge...
If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled, while the charge on the capacitor remains constant, then by what factor does the energy stored in the capacitor change? decreases by a factor of 2/3 increases by a factor of 6 decreases by a factor of 1/6 increases by a factor of 3/2 increases by a factor of 2
A parallel plate capacitor has square plates with sides of length 11 cm. The distance between the plates is 2 mm. The plates are charged up to 20volts. Part A What is the electric field between the plates? Express your answer using three significant figures. Electric field = N/C Part B What is the amount of charge on each plate? charge = C Part C What is the capacitance? Capacitance = μF Part D What is the energy stored by...
The plates of a parallel-plate capacitor are separated by 0.130 mm. If the material between the plates is air, what plate area is required to provide a capacitance of 3.50 pF? m2