62. A particle with a mass of 2.0x10 -5 kgand a charge of +2.0C is released in a

Question

62. A particle with a mass of 2.0x10-5kgand a charge of +2.0C is released in a (parallel-plate) uniformhorizontal electric field of 12 N/C. a. how far horizontally does the particle travel in 0.50s? b. What is the horizontal component of its velocity at thatpoint? c. If the plates are 5.0cm on each side, how much charge is oneach?
MUST GIVE ME THE ANSWER OR ELSE I"LL RATE NOT HELPFUL! 62. A particle with a mass of 2.0x10-5kgand a charge of +2.0C is released in a (parallel-plate) uniformhorizontal electric field of 12 N/C. a. how far horizontally does the particle travel in 0.50s? b. What is the horizontal component of its velocity at thatpoint? c. If the plates are 5.0cm on each side, how much charge is oneach?
MUST GIVE ME THE ANSWER OR ELSE I"LL RATE NOT HELPFUL!

Explanation / Answer

   Force onparticle   F   =   q *E   =   2.0 * 10-6 *12   =   2.4 *10-5   N    acceleration   a   =   F/ m   =   2.4 * 10-5 / 2.0*10-5   =   1.2   m/s2    a.   Distance travelled in tsec         s   =   u* t   +   (1/2) * a *t2    initialvelocity   u   =   0   =>   s   =   0 * 0.50   +   0.5 * 1.2 *0.52    =   0.150   m   =   15.0   cm    b.   Finalvelocity   v   =   u +a * t   =   0 + 1.2 *0.5   =   0.6   m/s    c.   Electric field betweenparallelplates         E   =   /0   =   Q /0 * A       A   =   Areaof plate   =   5 *5   =   25cm2   =   2.5 *10-3   m2          12   =   Q/ 8.85 * 10-12 * 2.5 * 10-3    Charge   Q   =   12* 8.85 * 10-12 * 2.5 *10-3   =   2.655 *10-13   C    Charge   Q   =   12* 8.85 * 10-12 * 2.5 *10-3   =   2.655 *10-13   C        

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