Question

A 2.3 mm -diameter sphere is charged to -4.5 nC . An electron fired directly at the sphere from far away comes to within 0.33 mm of the surface of the target before being reflected. a.What was the electron's initial speed? b.At what distance from the surface of the sphere is the electron's speed half of its initial value? c.What is the acceleration of the electron at its turning point?

Answer #1

R = radius of sphere = diameter/2 = 2.3/2 = 1.15 mm = 1.15 x
10^{-3} m

Q = charge on the sphere = - 4.5 x 10^{-9}C

q = charge on electron = - 1.6 x 10^{-19} C

v = initial speed = ?

m = mass of electron = 9.1 x 10^{-31} kg

r = closest distance of electron from center of sphere = R + 0.33 = 1.15 + 0.33 = 1.48 mm = 0.00148 m

using conservation of energy

kinetic energy = final electric potential energy

(0.5) m v^{2} = k Q q/r

(0.5) (9.1 x 10^{-31}) v^{2} = (9 x
10^{9}) (- 4.5 x 10^{-9}) (- 1.6 x
10^{-19})/(0.00148)

v = 9.81 x 10^{7} m/s

b)

v_{i} = initial speed = v = 9.81 x 10^{7}
m/s

v_{f} = final speed = v_{i}/2

r_{f} = final distance between electron and center of
sphere

Using conservation of energy

initial KE = final KE + final electric potential energy

(0.5) m v_{i}^{2} = (0.5) m
v_{f}^{2} + k Q q/r_{f}

(0.5) (9.1 x 10^{-31}) (9.81 x
10^{7})^{2} = (0.5) (9.1 x
10^{-31}) ((9.81 x
10^{7})/2)^{2} + (9 x 10^{9}) (- 4.5 x
10^{-9}) (- 1.6 x 10^{-19})/r_{f}

r_{f} = 1.97 x 10^{-3} m = 1.97 mm

d = distance from surface = r_{f} - R = 1.97 - 1.15 =
0.82 mm

c)

r = distance between electron and center at turning point = 1.48
mm = 1.48 x 10^{-3} m

F = force on the electron

F = k Qq/r^{2} = (9 x 10^{9}) (4.5 x
10^{-9}) (1.6 x 10^{-19})/(1.48 x
10^{-3})^{2} = 1.96 x 10^{-12} N

m = mass of electron = 9.1 x 10^{-31} kg

acceleration is given as

a = F/m = (1.96 x 10^{-12})/(9.1 x 10^{-31}) =
2.2 x 10^{18} m/s^{2}

A 1.00-mm-diameter glass sphere has a charge of + 1.40 nC.
What speed does an electron need to orbit the sphere 1.40 mm
above the surface?

In part (a) of the figure an electron is shot directly away from
a uniformly charged plastic sheet, at speed
vs = 2.20 × 105 m/s. The
sheet is nonconducting, flat, and very large. Part (b) of the
figure gives the electron's vertical velocity component v
versus time t until the return to the launch point. What
is the sheet's surface charge density?

1. What potential difference is needed to accelerate a He+ ion
(charge +e, mass 4u) from rest to a speed of 1.1×106 m/s
?
2. Two 2.00 cm × 2.00 cm plates that form a parallel-plate
capacitor are charged to ± 0.708 nC .
a) What is the electric field strength inside the capacitor if
the spacing between the plates is 1.30 mm ?
b)What is potential difference across the capacitor if the
spacing between the plates is 1.30 mm...

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