Question

An electron starts from red 4.00 an fron, the conter of a uniformly charged sphere of radus 2.00 em. " the sphere carnes a total charge of 1.00 x 10, С, how fast will the electron be moving when it reaches the surface of the sphere? m/s

Answer 1

Apply conservation of energy to solve this problem -

KEi + Ui = KEf + Uf

Initially the particle starts at rest, so initial KE is zero.

Ui - Uf = KEf

=> kqq' * (1/r1 - 1/r2) = (1/2)*m*v^2

where, r1 is the initial distance between the center of the sphere and the electron. r2 is the final position (right at the radius of the sphere). q is the charge of the sphere and q' is the charge of the electron. The entire right hand side should give you a positive number. K is a constant.

v = sqrt((2qq'/m)*(1/r1 -1/r2))

Now put the values -

q = 1.0 x 10^-9 C

q' = 1.6 x 10^-19 C

m = mass of electron = 9.109 x 10^-31 kg

r1 = 2.0 cm = 0.02 m

r2 = 4.0 cm = 0.04 m

So, v = sqrt[{(2 x 1.0 x 10^-9 x 1.6 x 10^-19) / (9.109 x 10^-31)}*(1/0.02 - 1/0.04)]

= sqrt[{3.2 x 10^-28} / (9.109 x 10^-31)}*25]

= sqrt[0.3513 x 10^3 x 25] = 93.71 m/s

Therefore, the velocity of the electron when it reaches at the surface of the sphere = 93.71 m/s (Answer)