A parallel-plate capacitor has square plates of side s = 2.30 cm and plate separation d = 3.00 mm. The capacitor is charged by a battery to a charge Q = 4.50 µc, after which the battery is disconnected. A porcelain dielectric (x = 6.5) is then inserted a distance y = 1.00 cm into the capacitor (see figure below). Hint: Consider the system as two capacitors connected in parallel. Dielectric K (a) What is the effective capacitance of this capacitor? pF (b) How much energy is stored in the capacitor?

In Figure 2, two parallel-plates capacitors (A and B) are connected in parallel acroSs a 240 V battery. Each plate has an area of 30 cm? and plates separation of 3.0 mm. Capacitor A is filled with air and capacitor B is filled with a dielectric of constant K= 2.3. (a) Explain a significance of having a dielectric between plates of capacitor B, (b) Find the magnitude of electric field within capacitors A and B (c) Find the potential energy stored in each capacitor. (d) Find the charge stored in each capacitor. B

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

A certain substance has a dielectric constant of 3.8 and a dielectric strength of 25 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 5.3 × 10-2 µ F and to ensure that the capacitor will be able to withstand a potential difference of 3.1 kV?

Review problem. A parallel-plate capacitor consists of square plates of edge length ℓ that are separated by a distance d, where d<<ℓ. A potential difference ΔV is maintained between the plates. A material of dielectric constant κ fills half the space between the plates. The dielectric slab is withdrawn from the capacitor as shown in the figure.   (a) Find the capacitance when the left edge of the dielectric is at a distance x from the center of the capacitor. (Use the following as necessary: ℓ, ε0, κ, x, and d.)        (b) If the dielectric is removed at a constant speed v, what is the magnitude of the current in the circuit, and its direction, as the dielectric is being withdrawn? (Use the following as necessary: ℓ, ε0, κ, ΔV, v, and d.)

Suppose you have the following system where there are two types of dielectric materials between capacitor plates. The dimension of the capacitor plates are w=15.0 cm, l= 23.0 cm and the separation between the plates are d= 10.0 cm. Dielectric material with dielectric constant Kị = 1.7 takes up the upper part of the capacitor with vertical distance h = 5.75 cm. And the remaining part is filled with material having dielectric constant k2 = 2.2. Ky We can replace two types of material with different dielectric constant by only one type of material with effective dielectric constant Keff. This way the capacitance of the whole system remains the same. Calculate the effective dielectric constant. effective dielectric constant Give your answer up to at least three significance digits.

A parallel plate capacitor has a charge of 5.50 x 10¬7C on one plate and -5.50 x 10-/ C on the other. The distance between the plates is increased by 46.0% while the charge on each plate stays the same. By how much does the energy stored in the capacitor increase? percentage

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.47 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 after AV f (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= = = (c) Determine the change in energy (in nJ) of the capacitor. AU = nJ AV f (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 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after…

A parallel plate capacitor with plate separation do is charged by connecting it to a battery. The energy stored by the capacitor is Ug. The battery is then disconnected and the plate separation is changed to d1 = 0.7do. If U1 is the energy stored by the capacitor when the plate separation is di, what is the ratio U1/Uo? O (a) 0.7 O (b) 0.837 O (c) 0.49 O (d) 2.04 O (e) 1.43 O (f) 1.2

A charged air-filled parallel plate capacitor of 22 nF has a energy density of 30x10-3 J/m³. If the distance between the plates is 200 µm, calculate the magnitude of the charge, in µC, of the capacitor.