62. A particle with a mass of 2.0x10 -5 kg and acharge of +2.0C is released in a

Question

62. A particle with a mass of 2.0x10-5kg and acharge of +2.0C is released in a (parallel-plate) uniformhorizontal electric field of 12 N/C. a. how far horizontally does the particle travel in0.50s? 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? 62. A particle with a mass of 2.0x10-5kg and acharge of +2.0C is released in a (parallel-plate) uniformhorizontal electric field of 12 N/C. a. how far horizontally does the particle travel in0.50s? 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?

Explanation / Answer

Initial speed (u) of the charged particle = 0 Mass (m) of the charged particle = 2.0x10-5kg Charge(q) of the charged particle =+2.0*10-6C Horizontal electric field (E) between the two charges = 12N/C (a) Horizontal distance (S) traveled by the charged particle inthe time 0.50s is                S = ut + (1/2)at2                   = 0 + (1/2)at2                   = (1/2)at2                    =(1/2)(Eq / m)t2 (b) Final speed (v) of the charged particle at this point is                v = u + at                   = 0 + (Eq / m)t (c) Charge on each plate of the capacitor is                Q = AEo                    = (L)2Eo Here E = 12 N/C           m =2.0x10-5kg            q= +2.0*10-6C            t= 0.50s            L= 5.0*10-2m          o =8.85*10-12C2/N.m2 Do all the calculation work. Do all the calculation work.

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