Question

Two very large charged parallel metal plates are 10.0cm apart and produce a uniform electric field of 2.80×10^6 N/C between them. A proton is fired perpendicular to these plates with an initial speed of 5.20 km /sec , starting at the middle of the negative plate and going toward the positive plate.

How much work has the electric field done on this proton by the time it reaches the positive plate?

Answer 1

Answer:

attached below

Explanation:

Answer 2

The work done by the electric field  on this proton by the time it reaches the positive plate will be  0.448 * $10^{-13}$ J

What is work done in an electric field ?

Electric field work is the work done by an electric field on a charged particle in its vicinity .

work done = force * displacement                         equation 1

given :

displacement = 10.0cm = 10 *  $10^{-2}$ m

to find = force =?

charge of proton = + 1.6 *$10^{-19}$C

electric field intensity = force / charge

force = electric field intensity  * charge

         = 2.80×$10^{6}$ N/C *  1.6 * $10^{-19}$ C

         = 4.48 *  $10^{-13}$ N

substituting the value of force and displacement  in equation 1

Work done = 4.48 * $10^{-13}$ N *  10 * $10^{-2}$ m

                  = 0.448 *  $10^{-13}$J

work done by the electric field  on this proton by the time it reaches the positive plate will be  0.448 * $10^{-13}$J

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