A particle (mass=6.7e27 kg, charge=3.2e19 C) moves along thepositive x axis with

Question

A particle (mass=6.7e27 kg, charge=3.2e19 C) moves along thepositive x axis with a speed of 4.8e5 m/s. It enters a region ofuniform electric field parallel to its motion and comes to restafter moving 2.0m into the field. What is the magnitude of theelectric field? I am unsure of which equation(s) to start with and was needinghelp with the problem. Need a response by today if possible beforemy test tomorrow. A particle (mass=6.7e27 kg, charge=3.2e19 C) moves along thepositive x axis with a speed of 4.8e5 m/s. It enters a region ofuniform electric field parallel to its motion and comes to restafter moving 2.0m into the field. What is the magnitude of theelectric field? I am unsure of which equation(s) to start with and was needinghelp with the problem. Need a response by today if possible beforemy test tomorrow.

Explanation / Answer

Thirdequation of motion relates final velocity (v), initial velocity(u), acceleration (a) and distance covered (s) as

   V2   =  u2   +   2 * a * s

   given   v   =   0

              u   =   4.8 * 105  m/s

              s   =   2.0   m

   02   =   (4.8 *10^5^2   +   2 * a * 2

   acceleration   a   =   -23.04 * 1010 / 4

                             =   - 5.76 * 1010  m/s2

   Here -ve sign indicates deacceleration and can bedropped off for calulations of magnitude.

   Force    F   =   m* a

                    =   6.7 * 10-27 * 5.76 * 1010

                    =   3.86 * 10-16   N

   Also   F   =   q * E

   Electric field    E  =   3.86 * 10-16 / 3.2 * 10-19

                                =   1.20 * 103   N/C

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