A 200µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change?
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- A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number Units (e) Number Units (f) Number UnitsQ.3: Four capacitors are arranged in the following configuration: the capacitances of capacitors are Ci= 3.0µF, C2= 2.0µF, C3= 1.8µF, C4= 3.0µF. The battery voltage is 12.0V. (a) Calculate the equivalent capacitance of the configuration. (b) How much energy is stored in this capacitor configuration? (c) How much energy is stored in capacitor C1? C1A large capacitance of 1.58 mF is needed for a certain application. (a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.19 µm of Teflon, which has a dielectric constant of 2.1. (b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is 60 ✕ 106 V/m (c) Find the maximum charge that can be stored. (d) Calculate the volume of Teflon alone in the capacitor. (units: m^3)
- In the figure below, capacitor 1 (C₁ Q₂ C₁ = = (a) Find the final charge on each capacitor after a long time has passed. с C 30.0 μF) initially has a potential difference of 55.0 V and capacitor 2 (C₂ = 5.10 μF) has none. The switches are then closed simultaneously. (b) Calculate the percentage of the initial stored energy that was lost when the switches were closed. %A 4.50 pF capacitor is connected in series with a 3.00 pF capacitor and a 350 V potential difference is applied across the pair. (a) What is the charge on each capacitor (in nC)? 4.50 pF саpасitor 6.3e-10 X nC 3.00 pF сараcitor 6.3е-10 X nC (b) What is the voltage across each capacitor (in V)? 4.50 pF сараcitor V 3.00 pF сараcitor VThe electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an 690-μF capacitor is 440 V. (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for 4.8 ms, find the effective power or "wattage" of the flash. (a) Number i Units (b) Number i Units >
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