Two parallel plates separated by 2.4 mm form a capacitor. The top plate has a voltage of -5 volts, and the bottom plate has a voltage of 1 volts. What is the electric field in the middle between the plates of the capacitor?
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- A mineral oil-filled capacitor is connected to a 9V battery. The capacitor is leaking oil while it remains connected to the battery. Consequently, we can expect this to happen as the capacitor continues leaking. O a. The electric field E between the plates will become stronger. O b. The electric flux density D in the capacitor will become weaker. c. The electric flux density D between the plates will become stronger. O d. The charge Q accumulated in the capacitor will remain the same.Two charged, square plates, separated by a distance of 13.1 cm and each with a side length of 20.3 cm, make up a parallel-plate capacitor. The electric field inside the plates is 1.40×103 N/C. If the charge is doubled, what would the electric field inside the plates become? 1.40E+03 N/C 700. N/C 2.80E+03 N/C 350. N/CAn electron is released from rest in a uniform electric field. If the electric field is 1.25 kN/C, at the end of 20 ns the electron's velocity will be approximately Group of answer choices 3.9 × 103 m/s. 4.4 × 106 m/s. 2.5 × 103 m/s. 3.0 × 108 m/s. 2.5 × 10–5 m/s.
- A parallel plate capacitor with a charge of +/-177nC has square plates length L=20.0cm across on each side. The plates are separated by a distance d=10.0cm. A positive +4.0microC point charge is held in place halfway between each plate on the right-most edge of the capacitor. While solving this problem ignore any distortion of the capacitor's electric field along its edges. A) What is the magnitude and direction of the capacitor's electric field at point P, which is midway between the plates and on the left-most edge of the capacitor B) What is the magnitude and direction of the electric field due to the +4.0 microC point charge at point P. C) What is the magnitude and direction of the net electric field at point PA 2.0 cm x 2.0 cm parallel-plate capacitor has a 1.0 mm spacing. The electric field strength inside the capacitor is 1.5×105 V/m. Potential difference across the capacitor =150 volts How much charge is on each plate?A simple and common technique for accelerating electrons is shown in the figure below, where a uniform electric field is created between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is 3.45 x 104 N/C. m/s² (b) Why will the electron not be pulled back to the positive plate once it moves through the hole? O The force of gravity is too strong. O There is no field outside the plates. O The other side of the positive plate also has a negative charge.
- A parallel-plate capacitor with rectangular plates has an electric field, E, between the plates. If the sides of the plates are doubled, the new electric field strength between the plates will beparallel-plate capacitors are widely used in electronics such as preventing direct current leakage. An electric field is generated between a pair of plates that has a surface charge density of 7.70 x 10^-6 C/m^2 The plates are seperated by 0.49 um. a) what is the magnitude of the electric field between the plates ? b) suppose the thickness of the plates doubled, would my answer increase, decrease, or remain the same?An air-filled capacitor is connected to a 9V battery. The air in the capacitor is carefully replaced with mineral oil while the capacitor remains connected to the battery. Consequently, we can expect this to happen. O a. The electric field E between the plates will become stronger. O b. The charge Q accumulated in the capacitor will remain the same. c. The electric flux density D in the capacitor will become weaker. O d. The electric flux density D between the plates will become stronger.
- What is the magnitude of the electric field and charge produced by a 25 cm x 15 cm capacitor separated by a 1.4 mm vacuum connected to a 12 V battery and c0 = 8.85 x 10^-12 C2/Nm2. a. The electric field is 8,571.4 V/m; the payload is 2.84 x 10^-10b. The electric field is 8,571.4 V/m; the payload is 2.84 x 10^-9c. The electric field is 8,751.4 V/m; the payload is 2.84 x 10^-10d. The electric field is 8,751.4 V/m; the payload is 2.84 x 10^-9A 5.30 cm by 2.40 cm parallel plate capacitor has the plates separated by a distance of 2.00 mm. When 4.00 × 10−11 C of charge is placed on this capacitor, what is the electric field between the plates?