A capacitor has a charge IQI = 3.0 x 103 C on each plate. It has a plate separation 4.0 x 103 m and capacitance C = 5.0 x 103 F. What is the magnitude of the electric field between the plates?
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- An air-filled parallel-plate capacitor has plates of area 2.00 cm2 separated by 1.70 mm. The capacitor is connected to a(n) 11.0 V battery. (a) Find the value of its capacitance. pF(b) What is the charge on the capacitor? pC(c) What is the magnitude of the uniform electric field between the plates? N/CA parallel-plate capacitor is formed from twO 9.0cm × 9.0 cm electrodes spaced 3.0 mm apart. The electric field strength inside the capacitor is 1.0 × 10° N/C.A 4 μF parallel-plate capacitor with a dielectric has a charge of 42 μC on each plate. The distance between the plates is 0.007 m and the dielectric constant is 153. Determine the area of each plate and the E-field between the plates. A = E =
- A parallel-plate capacitor has a plate area of 0.2m² and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10⁶ V/m between the plates, the magnitude of the charge on each plate should be: a. 9.0 x 10⁷ Cb. 2.0 x 10⁻⁶ Cc. 4.0 x 10⁻⁶ Cd. 7.1 x 10⁻⁶ CAn air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 2.00 mm. The capacitor is connected to a(n) 11.0 V battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/CTwo large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 195 V. An electron enters the region between them at a speed of 3.2 × 105 m/s by passing through a small hole in the negative plate and continues moving toward the positive plate. Assume the electric field between the plates is uniform. 1. What is the electron’s speed, in meters per second, when it is 0.1 cm from the negative plate? 2. What is the electron’s speed, in meters per second, when it is 0.5 cm from the negative plate? 3. What is the electron’s speed, in meters per second, when it is 1.0 cm from the negative plate? 4. What is the electron’s speed, in meters per second, when it is 1.5 cm from the negative plate? 5. What is the electron’s speed immediately before it strikes the positive plate?
- Two plates of area 865 x 104 m2 each are separated by a distance of 21.7 x 10-2 m. Calculate the capacitance C of the capacitor. C = FA shark is able to detect the presence of electric fields as small as 1.70 μV/m. To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 2.00-V battery. How far apart must the parallel plates be to have an electric field of 1.70 μV/m between the plates? need answer in kmA capacitor is composed of two metal plates. The two plates have the dimensions L = 0.11 m and W = 0.56 m. The plates have a distance between them of d = 0.1 m, and are parallel to each other. Part (a) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an expression for the magnitude of the electric field, |E|, halfway between the plates. Part (b) Input an expression for the magnitude of the electric field, |E2|, just in front of plate two. Part (c) If plate two has a total charge of q = -1 mC, what is its charge density, σ, in C/m2?
- 10. Ā = -2â + -3ŷ and B = -4â + -4ŷ. Calculate R = Ã+B. Calculate 0, the direction of R. Recall that 0 is defined as the angle with respect to the +x-axis. A. 49.4° B. 130.6° C. 229.4° D. 310.6°A charged conducting sphere with capacitance, C = 26.7 pF, produces an electric field that points radially outward. At a distance, d = 61.2 cm, from the center of the sphere, the strength of the electric field is E = 71.3 kV/m. (a) What is the radius of the charged sphere? R = m (b) Find the sphere's surface charge density. μC/m² J =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/C