An electron moving to the right at5.0% the speed of light enters auniform electr

Question

An electron moving to the right at5.0% the speed of light enters auniform electric field parallel to its direction of motion. If theelectron is to be brought to rest in the space of 5.7 cm, determine the following. this answer is 11207N/C. ok so i did it like this...E = ((1/2)*(9.109e-31 kg)*((0.05)*(2.998e+8m/sec))^2)/((1.602e-19 C)*(5.7e-2 m))
but, now itis not coming out right in my calc. can someone tell me what imdoing wrong? thanks!
An electron moving to the right at5.0% the speed of light enters auniform electric field parallel to its direction of motion. If theelectron is to be brought to rest in the space of 5.7 cm, determine the following. this answer is 11207N/C. ok so i did it like this...E = ((1/2)*(9.109e-31 kg)*((0.05)*(2.998e+8m/sec))^2)/((1.602e-19 C)*(5.7e-2 m))
but, now itis not coming out right in my calc. can someone tell me what imdoing wrong? thanks!

Explanation / Answer

   Use third equation ofmotion,   v2   =   u2   +   2* a * s    v   =   finalspeed   =   0,   u   =   initialspeed   =   5.0% *c   =   3.0 * 108 * 5 /100   =   1.5 *107   m/s    a   =   acceleration,   s   =   distance   =   5.7cm   =   0.057   m    =>   0   =   (1.5* 107)2   +   2* a * 0.057    a   =   - 1.97* 1015   m/s2    - ve sign is due to retardation and can bediscarded.    acceleration   a   =   force/ mass   =   q * E / m    =>   E   =   a* m / q   =   1.97 * 1015* 9.1 * 10-31 / 1.6 * 10-19       E      =      11.20   N/C       E      =      11.20   N/C

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