1. An ideal parallel- plate capacitor has a capacitance of C. If the area of the plates is halved and the distance between the plates is doubled, what is the new capacitance?
2. a proton has a. speed of 5.0x10^5 m/s at a point where the electrical potential is 500 V. It moves through a point where the electrical potential is 1000V. What is the speed at this second point? (e=1.60x10^-10C)

1. An ideal parallel- plate capacitor has a capacitance of C. If the area of the...
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
derive the capacitance of a parallel plate capacitor where each plate has area A and the plates are separated by a distance d. pleas show work step by step
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be -doubled -unchanged -quadrupled -quartered -halved
If the voltage between the plates of a parallel-plate capacitor is doubled, the capacitance of the capacitor:a) is halved.b) remains the same.c) is doubled.d) is tripled.e) quadruples.
A parallel-plate capacitor has capacitance 5.20 μF. The capacitor was origionaly connected to a 1.50 V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? Note: When disconnected, the charge on the capacitor must remain the same as when disconnected. A parallel-plate capacitor has capacitance 5.20 μF. The capacitor was origionaly connected to a 1.50 V battery? (c) The battery is subsequently reattached to the capacitor, but the plate...
A certain variable parallel plate capacitor has a capacitance of 33 μF. How much energy is stored in this capacitor if put at a potential difference of 7.5 V? The area of the plates is then doubled, and the plate separation is cut to 1/3 of its original value. It is then put at 7.5 V again. Now how much energy is stored in the capacitor?
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A and distance between plates d. Capacitor is connected to the battery of voltage V, fully charged and then disconnected. A slab of dielectric material with dielectric constant 4 is then inserted into capacitor, completely filling region between plates. Electric field between plates of capacitor: decreases twice stays the same increases twice decreases 4 times increases 4 times
A parallel-plate capacitor has a capacitance of 8 μF. Its capacitance if the plate separation is doubled is ? and if the plate area of the capacitor is doubled. The capacitance will be ?
Question 4 1 pts A parallel-plate capacitor of capacitance C has the area of plates A and distance between plates d. Capacitor is connected to the battery of voltage V, fully charged and stays connected. A slab of dielectric material with dielectric constant 4 is then inserted into capacitor, completely filling region between plates. Electric field between plates of capacitor: O stays the same O increases twice O decreases twice O decreases 4 times O increases 4 times
A parallel-plate capacitor has a capacitance of 140 pF, a plate area of 125 cm2, and a mica dielectric (κ = 5.4). At 55 V potential difference, calculate the following values. (a) E in the mica ???V/m (b) the magnitude of the free charge on the plates ???C (c) the magnitude of the induced surface charge on the mica ???C