An air filled capacitor of Co= 2.0 µF is fully charged with Qo=30.0 µC using an electromotive force of ɛ, as shown in the figure. The capacitor is then disconnected from the battery and filled with a dielectric material of constant K. If the electric potential across the capacitor falls to 1.9 V, then find K.

A spherical conductor whose center is at point O and radius R = 20 cm, is given a charge Q = + 100 μC a. Calculate the electric potential at points A and B if the distance OA = 30 cm and the distance OB = 45 cm. b. How much work is required to move an electron from point A to point B?

A positive charge is evenly distributed on the surface of a metal sphere centered at the origin of the coordinate system. The positive charge produces an electric field that points outwardly in the radial direction. A positive test charge is fixed at a distance R₁ from the center of the coordinate system, far from the surface of the sphere. The electric potential at R₁ is 10V. The positive test charge is then allowed to move freely until it reaches a distance R₂=2R₁ from the center of the coordinate system. What is the electric potential (in Volts) of the test charge at its new resting point? 10

A parallel-plate capacitor under the key of a keyboard has circular plates that are separated by 0.0500 mm and filled with a dielectric with k = 4.00. When the potential difference across the plates is 5.00 V , the magnitude of the charge on each plate is 8.897 pC. What is the radius of one of the plates? When the key is pressed, the distance between the plates decreases, increasing the capacitance to 1.85 pF. What is the separation between the plates when the key is pressed?

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.90 mm. A 15.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates.kV/m(b) Calculate the surface charge density.nC/m2(c) Calculate the capacitance.pF(d) Calculate the charge on each plate.pC

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 255 V. Assume a plate separation of d = 1.55 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q₁ = 364 PC after Qf = 29133 The charge will be the same before and after the capacitor is submerged in water. Apply the definition of a parallel-plate capacitor, and the relationship among C, Q, and AV. Don't forget to convert from cm to m as needed. pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= 1.142e-10 F X AV f = 255 This is the potential difference before the capacitor was submerged. When it is in the distilled water, the…

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 260 V. Assume a plate separation of d = 1.29 cm and a plate area of A = 25.0 cm2. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q; = 445.93 pC after Qf = 445.93 pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = 0.00000000013 AVf = 3.24 (c) Determine the change in energy (in nJ) of the capacitor. AU = -220 Write expressions for the initial and final energies, and take the difference. nJ (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 260 V potential difference.…

Suppose the parallel plates each have an area of 2000 cm? and are 1.00 cm apart. We connect the capacitor to a power (a) Vacuum (b) Dielectric supply, charge it to a potential difference Vo = 3.00 kV, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find the electric field E + Induced charges after the dielectric is inserted. 100,000 V/m 300,000 V/m 200,000 V/m 1000 V/m

A parallel plate capacitor initially charged with a battery that can produce a potential difference of 100 V is disconnected after being charged. A dielectric material with a dielectric constant of 4.6 and a thickness of 6 mm is placed symmetrically in the middle of the plates of this capacitor, which has a plate area of 0.18 m^2 and a plate separation of 1.8 cm. What is the capacitance of the capacitor after the dielectric material is inserted?A)3,2x10^-12FB)3x10^-6FC)1,12x10^-10FD)11x10^-6F

Consider the Earth and a cloud layer 500 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 4.0 km2 = 4000000 m2, what is the capacitance? F (b) If an electric field strength greater than 3.0 ✕ 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? C