A capacitor is made from two concentric spheres, one with radius 5.50 cm, the other with radius 7.80 cm. (a) What is the capacitance of this set of conductors (in pF)? pF (b) If the region between the conductors is filled with a material whose dielectric constant is 5.60, what is the capacitance of the system (in pF)? pF
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![A capacitor is made from two concentric spheres, one with radius 5.50 cm, the other with radius 7.80 cm.
(a) What is the capacitance of this set of conductors (in pF)?
pF
(b) If the region between the conductors is filled with a material whose dielectric constant is 5.60, what is the capacitance of the system (in pF)?
pF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2a435ab-e32d-4bb5-91c7-35592295b90b%2Ff5f421e5-2375-4eef-a536-2309fcc74ece%2Fkc91eem_processed.png&w=3840&q=75)
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- Four capacitors are connected as shown. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking ΔVab = 15.0 V.Two Capacitors are in series. C1=304 μF and C2=295 μF. What is the equivalent capacitance of the system measured in μF ? Enter a number and assume units of micro Farads.A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d. A parallel-plate capacitor is comprised of two square plates with sides of length ℓ and separation distance d. There are three dielectric materials in the space between the capacitor plates. A material with length ℓ⁄2, height d, and dielectric constant ?1 fills the left half of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?2 fills the top right quarter of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?3 fills the bottom right quarter of the space. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.)C = (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
- Three capacitors are connected in series as shown in the figure. The capacitances are C1 = 5.7 μF, and C2 = 9.8 μF, C3 is unknown, and the charge stored in each capacitor is Q = 8.5 μC. a) Express the capacitance C of a capacitor in terms of charge Q and voltage ΔV on it. Part (b) Apply the above formula to capacitor C1 to find an expression for the potential difference ΔV12 across it. Part (c) Express the potential difference ΔV12 through potentials V1 and V2 where V1 and V2 are the potentials measured in the wires 1 and 2, respectively, relative to the negative side of the battery. Part (d) Calculate V2 in V given V1 = 9 V. Part (e) Repeat the above procedure for capacitor C2 and calculate the potential at point 3, V3 in V.please answer vCapacitance Problem 7: Consider a spherical metal shell of radius R in empty space. Part (b) Calculate the capacitance, in picofarads, of such a conductor with a radius of R = 0.058 m.
- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius ry has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = aer/a0 + B/r + bo where alpha (a), beta (8), an and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vah between the two conductors. The potential difference is related to the electric field by: Vah = Edr= - Fdr Calculating the antiderivative or indefinite integral, Vab = (-aager/a0 +8 + bo By definition, the capacitance Cis related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C= Q/( (e"b/a0 .eralao) + B In( )+ bo ( ))A system of capacitors is shown below. What is the equivalent capacitance of the system is the individual capacitors are; Express the answer in microFarads. For example, if the answer came out to be 4.98 X 10-7 F, this is the same as 49.8 microFarads. You would then answer with the numeric value of 4.98. C1 = 5.5 x 10-6 F C2 = 5.7 x 10-6 F C3 = 8 x 10-6 F C4 = 4.5 x 10-6 F C5 = 2.1 x 10-6 FA capacitor is attached to a battery. It has initial capacitance Co, voltage Vo, charge Qo, and energy Uo. Let's put some dielectric material of dielectric constant K in there in two different ways. For each case, figure out the following for the resulting "new" capacitor: C, V, Q, and U, in terms of the old value and K. The capacitor is left attached to the battery. A slab of dielectric material with K=3 is inserted into the capacitor without otherwise disturbing it. In terms of the initial values, what are the new values? The capacitor is disconnected from the battery after it is charged up to Qo. Then, a slab of dielectric material with K=3 is inserted into the capacitor without anybody touching the plates or otherwise disturbing the capacitor. In terms of the initial values, what are the new values?
- A system of capacitors is shown below. What is the equivalent capacitance of the system is the individual capacitors are; Express the answer in microFarads. For example, if the answer came out to be 4.98 X 10-7 F, this is the same as 49.8 microFarads. You would then answer with the numeric value of 4.98. C1 = 6.7 x 10-6 F C2 = 4.2 x 10-6 F C3 = 2.1 x 10-6 F C4 = 2.6 x 10-6 F C5 = 6.3 x 10-6 FThe dielectric constant of a solid is 2000. It is placed between the plates of a capacitor that are 1 μm apart and with an applied voltage of 0.10 V. Calculate the capacitance per square millimetre of this device and the local electric field acting on an atom in the dielectric.The plates of a spherical capacitor have radii 35.7 mm and 37.8 mm. (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance? (a) Number i Units (b) Number i Units