fd C1 K fd K d (1 – f)d (1 – f)d C2 a b Figure 25.21 (Example 25.8) (a) A parallel-plate capacitor of plate separation d partially filled with a dielectric of thickness fd. (b) The equivalent circuit of the capacitor consists of two capacitors connected in series.

Capacitor A is a standard parallel-plate capacitor with no dielectric. It was charged up (though is currently not attached to anything) and currently has charge of Qo, a voltage of Vo, a capacitance of Co, and a potential energy of PE,. A dielectric with K = 41 is to be inserted into capacitor A. Determine what the capacitance, charge, voltage, & potential energy of capacitor A will be once the dielectric is fully inserted. C = C,x 41 Q = Q,x 1 V = V,x PE = PE,X

What is the electric charge, in the unit of μC, stored in a capacitor with C = 42 μF if the voltage across the capacitor is 1.6 V

A system of capacitors is shown below. What is the equivalent capacitance of the system if 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 numeric value of 4.98. C1 = 6.6 x 10-6 F C2 = 5.9 x 10-6 F C3 = 3.4 x 10-6 F C4 = 4.4 x 10-6 F C5 = 8.1 x 10-6 F

It turns out that the dielectric k constant increases the capacitance by a factor of k.  That is, Cnew=kCold.  In this case, with the capacitor not connected to a battery, by what factor does the voltage difference across the capacitor change?

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.

A 10 cm2 parallel plate capacitor has its plates separated by 5 x 10-4 m thick paper. Find the capacitance of the capacitor. (κpapel = 3,7)

A parrallel plate capacitor consisting of two 30.0 cm x 30.0 cm square conductors are separated by a 3.00 mm piece of mica (k=5.00) is charged by a connecting it to an 18.0 V battery. What is the capacitance of this capacitor? What is the charge on the capacitor? If the mica is removed and replaced with glass (k=7.00), what is the potential difference of the plates now? What is the charge on the capacitor now with the glass?

A capacitor has a voltage difference between its terminals of 1.3 V. The amount of charge stored in the capacitor is 16 uC. What is the capacitance of the capacitor?

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 F