Question

80. A thin conducting plate 1.0 m on the side is given a charge of $-2.0 \times 10^{-6}$ C. An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron? 

          80. A thin conducting plate 1.0 m on the side is given a charge of $-2.0 \times 10^{-6}$ C. An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron?
80. A thin conducting plate 1.0 m on the side is given a charge of -2.0 × 10^-6 C. An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A thin conducting plate 1.0m on the side is given a charge of -2.0 imes 10^(-6)C. An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron? 80. A thin conducting plate 1.0 m on the side is given a charge of --2.0 10-6 C. An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron?
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Transcript

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00:02 Hello and welcome to this video solution of numerate here we are given a thin conducting plate having one meter on each side right is given a charge q which is negative 2 times of 10 to the power of minus 6 coulomb right and there is the electron that is placed at a distance of one centimeter of the center of the plate right based on this you have to calculate the acceleration of this electron, right? so first of all what we need to do is we need to calculate the electric field that is produced by this parallel plate, right? so it's nothing but q over 2 or maybe the expression is sigma which is the surface charge density over 2 epsilon naught, right? so there's a value of electric field and this sigma is nothing but the charge over the area right of the plate so this is the electric field you will be having next what we have is a force on the electron, right? so f will be equal to the charge of the electron e times this electric field, right? so it's q e over 2 epsilon naught a right? so there's a net force the acceleration is equal to the force of the electron over the mass of the electron right which is q e over 2 epsilon naught a times m right so we now we plug in all these values so it's 2 times of 10 to the power of minus 6 i'm neglecting the signs you need to only calculate the magnitude right 1 .6 times of 10 to the power of minus 19 over 2 times of 8 .85 times of 10 to the power of minus 12 this is epsilon naught area is 1 square and the mass of the electron is 9 .1 times of 10 to the power of minus 31 kgs, right? this will be in meters per second...
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