Each of the electrons in a particle beam has a kinetic...

Each of the electrons in a particle beam has a kinetic energy of 1.60 × 10^(-17) J. (a) What is the magnitude of the uniform electric field (pointing in the direction of the electrons' movement) that will stop these electrons in a distance of 10.0 cm? (b) How long will it take to stop the electrons? (c) After the electrons stop, what will they do? Explain.

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00:01 Hello students solution for the given problem is solution solution for part a of the given problem we know that work done w is equal to f multiply by s that is equal to q e multiply by s equation first from work energy turin we know that w is equal to del change in kinetic energy energy that is equal to w is equal to negative k i where k f is equal to zero so q e dot s is equal to negative k i by substituting the values in equation second q which is negative 1 .6 multiply by 10 power negative 19 multiply by e multiply by 0 .0 .0 .0 0 .07 value of s is equal to initial kinetic energy which is negative 1 .60 multiplied by 10 power negative 17 so that gives us the value of e as equal to 142 at 0 .5n over c now solution for part b of the given problem is from newton's second law we know that f is equal to m a and m a is equal to negative qe or a is equal to negative qe over m equation third but a is equal to v initial minus final over t that is equal to v minus v0 over t is equal to v minus v0 qe over m where v is equal to 0 so negative v0 over t is equal to negative qe over m and initial kinetic energy k i is equal to 1 over 2 m v0 square so v0 v0 is equal to under root 2k i over m so t will be equal to under root 2 k i m multiply by m q a by by substituting the values in equation 4th t is equal to under root 2 multiply by initial kinetic energy which is 1 .60 multiply by 10 plus power negative 17 divide by 1.

03:39 6 multiplied by 10 with power negative 9 multiplied by 414 to add 4 .5 multiply by 9 .11 multiply by 10 with power negative 31 so that gives us the value of t is equal to 2 .36 multiplied by 10 plus power negative at second and solution for part c of the given problem is we know that a is equal to negative q e over m that is equal to negative 1 .6 multiply by 10 power negative 19 multiply 1 428 .5 over 9 .11 multiply by 10 plus power negative 31 so that gives us the value of a a is equal to 2 .50 multiply by 10 power 14 meter per second square so direction opposite to field...

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