A 2.6 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm of the surface of the target before being reflected. What was the electron's initial speed? At what distance from the surface of the sphere is the electron's speed half of its initial value? What is the acceleration of the electron at its turning point?
A) 9.3x10^7 m/s but I cannot figure out B or C
Answer
a) electron's initial kinetic energy = electron's electric potential energy before being reflected
1/2mv^2 = K q e/r
q=-4.6e-9 C ; e= charge of electron= 1.602e-19 C
(9.11e-31 kg)(v^2)/2 = (9e9 Nm2/C2) (-4.6e-9 C)(-1.602e-19 C)/(2.6e-3 / 2 + 3.9e-4 m)]
v = 9.3e7 m/s
b) Let x be the distance from the surface of the sphere where the electron's speed half of its initial value
electron's initial kinetic energy = (kinetic energy of electron at x ) + (electric potential energy of electron)
1/2mv^2 =1/2 m(v/2)^2+ K q e /r
1/2mv^2 =1/2 mv^2 /4+ K q e /r
1/2mv^2 -1/2 mv^2 /4 = K q e /r
3/8mv^2 = Kqe/r
3(9.11e-31 kg)(9.3e7 m/s)^2 / 8 = (9e9 Nm2/C2) (-4.6e-9 C)(-1.602e-19 C)/(2.6e-3 / 2 m + x)]
x = 9.5e-4 m = 0.95 mm
c) for acceleartion we use F= ma
ma = K qe/r2
(9.11e-31 kg)(a) = (9e9 Nm2/C2)(-4.6e-9 C)(-1.602e-19 C) / (2.6e-3 / 2 + 3.7e-4 m)^2]
a = 2.5e18 m/s^2
