Download Quiz 3 Solution

Physics 7D Quiz 3
Name: Jordan Smolinsky
Two flat plates with charge density σ = +5nC/m2 and −σ = −5nC/m2 , respectively, lie
parallel to the xy plane, separated by a distance d = 2 cm. An electron (charge −1.6 × 10−19
C, mass 9.11 × 10−31 kg) enters the plates at height d/2 = 1 cm moving in the positive
ŷ direction with speed 106 m/s. What is the electron’s speed when it hits the positively
charged plate?
Solution
This problem can be approached in two ways: either by finding the force exerted by the
plates’ electric field on the electron or by using conservation of energy. Since you have just
covered potentials the preferred solution, which I will present here, is the latter. First we
~ = σ/0 ẑ
want to find the potential between the plates. Because the electric field is constant E
the potential difference between any two heights is simple to calculate.
~ · ~l =
∆V = E
σ
∆z
0
(1)
Therefore the potential difference between where the electron enters the plates and where
it hits the positively charged plate is given by
∆V =
σd
20
(2)
eσd
20
(3)
and the potential energy lost is then
∆U = −
To find the final velocity we use conservation of energy. The final velocity is related to
the initial velocity and the change in potential energy by
1
eσd
1
me vi2 +
= me vf2
2
20
2
r
eσd
vf = vi2 +
me 0
Plugging in our numbers we obtain 1.73 × 106 m/s.
1
(4)