potential and potential energy in a uniform electric field?

lina

2007-05-20 8:36 am

1.suppose a 12-V car battery is used as a source of energy to accelerate particles.find the speeds that would be at-tained by (a)an electron;and(b)a proton.assume the particles start at rest.
2.starting at a point 1 m away from a 2nC charge,how far in the radial direction are the points at which the poten-tial is (a)1 V higher;(b)1 V lower?

回答 (1)

天同

2007-05-21 2:55 am

✔ 最佳答案

1. Gain in kinetic enrgy = q.V
where q is the charge(in Coulombs) and V is the potential diference (p.d) in volts
thus (1/2)m.v^2 = qV
where m is the mass of the particle and v is its speed
v^2 = 2qV/m
for electron and proton, q = e=1.6x10^-19 C
v^2 = 2 x1.6x10^-19x12/m
substitute the masses of electron and proton respectively into m, and solve for v
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2. Potential V at 1m from the charge,
V = kq/r
where q is the charge(=2x10^-9 C), r is the distance from the charge(=1 m), and k is a constant(=9x10^9 N.m^2/C^2)
thus V = 9x10^9x2x10^-9/1 volts = 18 volts
At point of potential 1 volt higher, i.e. 19 volts, let the distance (from the charge) be r1
thus 19 = 9x10^9x2x10^-9/r1
solve for r1 gives r1 = 18/19 m
Likewise, for distance r2 where the potential is 1 volt lower (i.e. 17 volts),
r2 = 18/17 m

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