Two 2.0-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0x105 V/m. What is the voltage across the capacitor? Express your answer with the appropriate units. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. HẢ ? V= Value Units Submit Request Answer Part B How much charge is on each disk? Express your answers in coulombs separated by a comma. ? q1, 92 = C Submit Request Answer

Two 2.0-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0x105 V/m. What is the voltage across the capacitor? Express your answer with the appropriate units. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. HẢ ? V= Value Units Submit Request Answer Part B How much charge is on each disk? Express your answers in coulombs separated by a comma. ? q1, 92 = C Submit Request Answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A parallel-plate capacitor has the capacitance of C = 10 µF when the distance of the two plates is d = 5 cm. A battery with the emf = 3 V is used to charge the capacitor. %3D 1. How much charge can be put on the parallel-plate capacitor? (in the unit of pC) Q = Submit Answer Tries 0/2 Now, consider the following two cases separately: Case 1: if the emf stays connected, but the distance of the parallel plates is changed to d = 80 cm. 2. What is the new capacitance C1 and how much is the charge Q1in the capacitor? C1 = µF %D Q1 = %3D Submit Answer Tries 0/2 Case 2: After the initial charging, the emf is removed. Therefore, the amount of charge you calculated from the question #1 remains in the capacitor. Now, again, the distance of the parallel plates is changed to d1 = 80 cm. 3. What is the potential difference V1 across the capacitor? V1 = %3D
The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by the distance between them. When a kV voltage of 135. V is put across the plates of such a capacitor an electric field strength of 1.1 is measured. cm Write an equation that will let you calculate the distance d between the plates. Your equation should contain only symbols. Be sure you define each symbol. Your equation: d=0 Definitions of your symbols: kV = 1.1 cm 135. V Explanation Check 20 E A S tv Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessib
B. Problem Solving. Solve the ff. problems by providing a solution. Round-off your answer to the nearest hundredths (with 2 decimal places). 1. A parallel-plate capacitor has square plates 3.5 cm on a side, separated by 0.50 mm. The capacitor is charged to 10 V, then disconnected from the charging power supply. (a) Calculate the capacitance of this capacitor. (b) What is the total charge on each plate?
Three capacitors are connected as shown, where C = 0.045 F. A. Input an expression for the equivalent capacitance between points a and b, in terms of C. B. What is the capacitance, in farads? C. If the capacitors are charged with a ΔV = 10 V source, how much energy will the circuit store, in joules?
1. You have five 10.0 F capacitors. Show all the possible connections for the five capacitors to produce a total capacitance of 50.0 F. 2. Suppose the space between the plates of the capacitor in Sample Problem 2.6 item number 2 is filled with equal thicknesses of the same dielectrics but arranged as shown. What is its capacitance? 3. The capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor.
1. You have five 10.0 F capacitors. show all the possible connections for the five capacitors to produce a total capacitance of 50.0 F. 2. Suppose the space between the plates of the capacitor in sample problems 2.6 item number 2 is filled with equal thickness of the same dielectrics but arranged as shown. What is its capacitance?
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 parallel plate capacitor in air is constructed with two 13 cm x 13 cm square conductors separated by 3 mm. a) Determine the value of the capacitance of this parallel plate capacitor. b) This capacitor is placed across a 12 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.
A parallel plate capacitor is designed as shown in Fig.The capacitor has a cross-sectional area of A = (1,2 × 1,2) cm2, the distance between the plates d= 2 mm, where the dielectric constant k1 = 4, k2 = 6, and k3 = 2.a. Describe the capacitor combination (series, parallel) formed byall three dielectric materials.b. Calculate the total capacitance of the capacitor?c. If the capacitor is connected with a potential difference V = 5 voltsthen determine the charge and energy stored in the capacitorthe?
PLEASE ANSWER BELOW. IT IS A RESEARCH HYPOTHESIS 1. A capacitor was made of aluminum foil strips which were separated by Mylar films. The capacitor was subjected to too much voltage and produced holes in Mylar films. The capacitance was still found to have the same magnitude but the voltage was lesser. Why do you think this happened? 2. Is there any kind of material that when inserted between parallel plate capacitors would reduce its capacitance?
The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by the distance between them. When a kV voltage of 129. V is put across the plates of such a capacitor an electric field strength of 3.7- is measured. cm Write an equation that will let you calculate the distance d between the plates. Your equation should contain only symbols. Be sure you define each symbol. Your equation: d = Definitions of your symbols: kV = 3.7 cm = 129. V 00 X E Δ
P1. A plane parallel capacitor, whose plates are eadh of area A and separation d, is ch arged with a bat tery of E MF V. How much work is done if a large met allic slab with thickness d/3 is in serted in side the capacitor? The slab is parallel with the plates. The capacitor is connected to the battery while we move the metallic slab, so the potential of the capacitor is fixed. metal d/3 d A