An electron is fired through a small hole in the positive plate of a parallel-plate capacitor at a speed of 1.1 x 107 m/s. The capacitor plates, which are in a vacuum chamber, are 1.8-cm-diameter disks spaced 3.0 mm apart. The electron travels 2.0 mm before being turned back. What is the capacitor's charge? Express your answer with the appropriate units. μA Q= Value Units ?

An electron is fired through a small hole in the positive plate of a parallel-plate capacitor at a speed of 1.1 x 107 m/s. The capacitor plates, which are in a vacuum chamber, are 1.8-cm-diameter disks spaced 3.0 mm apart. The electron travels 2.0 mm before being turned back. What is the capacitor's charge? Express your answer with the appropriate units. μA Q= Value Units ?

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter7: Electric Potential

Section: Chapter Questions

Problem 99AP: Two parallel conducting plates, each of cross-sectional area 400 cm2, are 2.0 cm apart and...

See similar textbooks

Similar questions

Two parallel conducting plates, each of cross-sectional area 400 cm2, are 2.0 cm apart and uncharged. If 1.01012 electrons are transferred from one plate to the other, (a) what is the potential difference between the plates? (b) What is the potential difference between the positive plate and a point 1.25 cm from it that is between the plates?
In nuclear fission, a nucleus splits roughly in half, (a) What is the potential 2.001014 in from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?
The practical limit to all electric field in air is about 3.00106 N/C. Above this strength, sparking rakes place because air begins to ionize, (a) At this electric field strength, how far would a proton travel before hitting the speed of light (ignore relativistic effects)? (b) Is it practical to leave air in particle accelerators?
(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is 3.00106 V/m, if the plates are separated by 2.00 mm and a potential difference of 5.010V is applied? (b) How close together can the plates be with this applied voltage?
(a) At what speed will a proton move in a circular path of the same radius as the electron in the previous exercise? (b) What would the radius of the path be if tlie proton had the same speed as the election? (c) What would the radius be if the proton had tlie same kinetic energy' as die electron? (d) The same momentum?
Problem 2: Consider the parallel-plate capacitor shown in the figure. The plate separation is 5.9 mm and the the electric field inside is 24 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v₁. Randomized Variables d = 5.9 mm E = 24 N/C + + + + + + + + d Otheexpertta.com Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? Part (b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case? select part Vf,b=1
... Path of trajectory AV w An electron is fired at a speed v¡ = 3.4 x 106 m/s and at an angle 0; = 30.5° between two parallel conducting plates as shown in the figure. If s = 1.5 mm and the voltage difference between the plates is AV = 98 V, determine how close, w, the electron will get to the %3D %3D bottom plate. Put your answer in meters and include at 6 decimal places in your answer. Do not include units. The x-axis of the coordinate system is in the middle of the parallel plate capacitor. Round your answer to 6 decimal places.
Two square, conducting plates each contain charge of magnitude Q = 35.4 nC. The plates have side length l = 20 cm and are separated by a distance d = 40 mm. Electrons are fired at speed v throgh a tiny hole in the positive plate toward the negative plate. (Remeber that E = Q/Ep0A for parallel plates) a. what is the potential difference between the plates b. if the electrons are stopped halfway between the plates, what is their initial speed, v? c. what is teh minimum speed the electrons must have if they are to make it all the way to the negative plate?
Problem 2: Consider the parallel-plate capacitor shown in the figure. The plate separation is 4.7 mm and the the electric field inside is 26 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v₁. Randomized Variables d=4.7 mm E = 26 N/C Vi,a + + + + + + + + Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? d
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
A parallel-plate capacitor is constructed with two disks spaced 2.00 mm apart. It is charged to a potential difference of 500 V. A proton is shot through a small hole in the negative plate (point A) with a speed of 2.0 x 105 m/s. a. Use the law of conservation of energy and determine the potential at the point where the speed of the proton is zero (point B, shown in the diagram) b. How far is the point B from the negative plate of the capacitor? OV Before: = 0 mm E 500 V After: B V=0 m/s X d = 2.00 mm
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=