An ionized nitrogen molecule (N2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.796 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of a nitrogen molecule is 4.65 ✕ 10−26 kg and the fundamental charge is  e = 1.60 ✕ 10−19 C.)   (a) the x-component of the electric field (in V/m)    (b) the potential difference between points A and B (in V)

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.

A uniform electric field of magnitude 395 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 2.90 cm. (a) What is the work done by the field on the electron? 1.83e-16 Your response is off by a multiple of teh. J (b) What is the change in potential energy associated with the electron? -1.83e-16 Your response is off by a multiple of ten. J (c) What is the velocity of the electron? 2.00e-24 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s Select v magnitude direction Need Help? Read It

The electric field in a certain region is given by the equation E = (axn - b) i where a = 11.14 N/(Cm6), b = 9.02 N/C and n = 6 Calculate the potential difference in volts from point x1 = 0.77m to point x2 = 1.88m

Oppositely charged parallel plates are separated by 6.30 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.80 mm from the positive plate?

An electron with charge e and mass m is accelerated from rest through a potential V over a distance d. The electric field E in the region is constant. Which of the following expressions can be used to correctly calculate the speed v of the electron as it leaves region?     eV = 1/2 mv2     eV = mv     Ed = m2v2     v2 = 2Ee/d

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.56 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.   ????V/m

Problem 2: A point mass of charge q > 0 moves at 35° to a uniform electric field E for a total distance d. Part (a) Select a diagram that follows from the problem description. Part (b) The charge q travels from point y, to point y2, where the total traveled distance is d. Find the potential difference that the charge traverses. When you enter your answer you can assume that y, = 0 V2 - V = cos(a) cos(o) cos(e) 7 8 9 HOME sin(a) sin(@) sin(e) 5 6 3 d E - END h Vol BACKSPACE DEL CLEAR

Oppositely charged parallel plates are separated by 5.35 mm. A potential difference of 500 V exists between the plates. (A) what is the magnitude of the electric field between the plates? (B) how much work does the electric field do moving an electron from the negative plate to the positive plate?

Starting from rest, a proton was accelerated by an electric field. After a distance of 1.35 m the proton reaches a speed of 1.97 x 105 m/s (m, = 1.67 x 10-27 kg, q,= 1.60 x 10-19 C). a) What is the final kinetic energy of the proton? b) What is the potential energy difference of the proton? c) What is the magnitude of the electric field?