A nervous physicist worries that the two metal shelves of his wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. What is the capacitance of the empty shelves if they have area 0.375 m2 and are 0.175 m apart? What is the voltage between the two shelves if opposite charges of magnitude 3.25×10−9 C are placed on them? Calculate the energy stored by having this charge on the two shelves.

A parallel-plate capacitor with area 0.220 m² and plate separation of 2.00 mm is connected to a 7.40-V battery. (a) What is the capacitance? F (b) How much charge is stored on the plates? C (c) What is the electric field between the plates? N/C (d) Find the magnitude of the charge density on each plate. C/m2 (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?

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?

A 12-volt battery is attached to two parallel metal plates. The dimensions of each plate are 20 cm by 40 cm. The plates are separated by a piece of glass, with dielectric constant k = 11.3, which has thickness 5.19 mm. The plates have equal but opposite charges +Q; calculate the value of Q. (in C) OA: 3.343x10-9 B: 4.446x10-9 OC: 5.913x10-9 OD: 7.864×10-9 E: 1.046×10-8 OF: 1.391x10-8 OG: 1.850x10-8 OH: 2.461x10-8

The potential difference between a storm cloud and the ground is 2.47 x 10° V. If a bolt carrying 5 C falls from a cloud to Earth, what is the magnitude of the change of potential energy of the charge? Answer in units of J.

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An engineer connects a 24.0 V battery to a capacitor of unknown capacitance. The result is that 52.0 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?

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 small object with a mass of 3.2 mg carries a charge of 29.6 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 5.94 cm. If the thread makes an angle of 10.2° with the vertical, what is the potential difference in volts between the plates?

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.…