A proton has an initial speed of 3.0×105 m/s . Part A What potential difference

Question

A proton has an initial speed of 3.0×105 m/s .

Part A
What potential difference is required to bring the proton to rest?
Express your answer using two significant figures.
  
  
V =    V

Part B
What potential difference is required to reduce the initial speed of the proton by a factor of 2?
Express your answer using two significant figures.

V =    V
  

Part C
What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?
Express your answer using two significant figures.

V =    V

Explanation / Answer

(a)formula for potential due to point charge is

V = kq/r

from the conservation of energy

W = q del V

del KE = q del V

KEi- KEf = q del V

since proton final velocity is zero KE f= 0

1/2 m v^2-0 = q del V

del V = 0.5(1.6726216 * 10-27 kg)(3.0 * 105 m/s)2/(1.6021765 * 10-19 C)

=469.77 V

for two significat figure 470 V

(b)

1/2 m vi^2-1/2 mvif^2 = q del V

0.5 m ( vi^2- vf^2) = q del V

del V = 0.5(1.6726216 * 10-27 kg)(3.0 * 105 m/s)^2- (1.5 * 10^5)2/(1.6021765 * 10-19 C)

=351.82V for two significant digits 350 V

(c)

kf= ki/2

Apply conservation energy

ki+ q Vi = kf+ q V f

ki- ki/2 = q del V

ki/2= q del V

del V= ki/ 2q

= 0.5(1.6726216 * 10-27 kg)(3.0 * 105 m/s)2/2(1.6021765 * 10-19 C)

=235.12 V

for two significant digits 240 V

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