A 7.5 nF capacitor has 8.5 μJ of energy stored in its electric field. If the capacitor fully discharges in 0.1ms how much charge in nC will be released? Round your answer to the nearest integer.
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- A capacitor has a capacitance of 6μF with air as thedielectric. A battery charges the capacitor to 400 V and isthen disconnected. What is the new voltage if a sheet ofmica (K = 5) is inserted? What is new capacitance C ?a. Find the capacitance C and the charge Q if connectedto 200-V battery. Assume the dielectric constant is K =5.0.b. A capacitor has a capacitance of 6μF with air as thedielectric. A battery charges the capacitor to 400 V and isthen disconnected. What is the new voltage if a sheet ofmica (K = 5) is inserted? What is new capacitance C ?c. Find the equivalent capacitance of a circuit in which a6-μF capacitor is connected in series with two parallelcapacitors whose capacitances are 5 and 4 μF.A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cmcm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 10.0 cmcm. The capacitance is 38.5 pFpF. A. Calculate the outer radius of the hollow tube.Express your answer in centimeters. B.When the capacitor is charged to 125 V, what is the charge per unit length λ on the capacitor?Express your answer in coulombs per meter.
- A parallel plate capacitor in air is constructed with two 32 cm x 32 cm square conductors separated by 4 mm. a) Determine the value of the capacitance of this parallel plate capacitor. b) This capacitor is placed across a 22 V battery and allowed to fully charge. What is the value of this charge? To continue, please enter your answer in b) in units of nC. Round your answer to 1 decimal place.Useful Constants: k = 9.00 × 10º Nm² C2 €0 = 8.85 × 10–12 C² Nm2 e = 1.6 × 10¬19C %3D mẹ = 9.11 x 10~31 kg %3D 27 kg = 1.67 × 10 mp mn = 1.68 × 10-27kgA parallel plate capacitor has square plates 6cm on a side, separated by 0.3mm. The capacitor is charged to 3V,and then disconnected from charging power supply. 1.1 What is the total charge? 1.2 What is the charge density on each plate?
- A laboratory capacitor consists of two parallel conducting plates, each with area of 200 cm² separated by a 0.40-cm air gap. a. Compute its capacitance b. If the capacitor is connected across a 500-V source, find the charge on it, the energy stored in it, and the value of E between the platesA capacitor of capacitance 9.5 milli-farads has a charge of 2.4 milli-coulombs. What is the voltage across the capacitor plates? Answer in volts.A capacitor is formed from a hollow copper sphere of inner radius 36.3 cm and outer radius 45.4 cm. Calculate the capacitance C of the sphere in microfarads. C = HF
- It is the gap between the conductors in a capacitor consisting of two concentric spherical conductors. The radius of the inner conductor is 10 cm and the capacitance of the capacitor is 116 pF. What is the radius of the outer conductive sphere?The unit of the electric "coulomb" constant k is (N.m²/C² ) An equivalent unit would be: O a Watt.s?/c².m? Ob Watt.m.s²/C? Oc Watt.s/C2m Od Watt.m.s/C² De Watt.m2. s/C?Suppose you have two parallel conducting plates that are separated by 2.2 mm. a. What will the electric field strength between the plates be (in N/C) if they have a potential difference of 4.6 × 103 V? E = b. The electric breakdown strength for a particular medium, also called the dielectric strength, is the point at which electrons bound to the molecules of the medium begin to be stripped off due to the large electric field. How close together must the plates be with this applied voltage in order to achieve breakdown strength for air (3.0 × 106 V/m) in mm? dbreakdown=