Question

A small charged bead has a mass of 3.8 g . It is held in a uniform electric field vec E = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 26 m/s^2 . Part A What is the charge on the bead? Express your answer to two significant figures and include the appropriate units. q = 6.42.10^-7 C

Answer 1

Here ,

m = 3.8 Kg

E = 2*10^5 N/C

a = 26 m/s^2

as Second law of motion

Fnet = ma

qE - mg = ma

q*2*10^5 = 3.8*(9.8 + 26)

q = 6.8 *10^-4 C

the charge on the bead is 6.8 *10^-4 C

Answer 2

Fnet = qE - m*g

m*a = q*E - m*g

==> q = m*(a+g)/E

= 3.8*10^-3*(26+9.8)/200000

= 6.8*10^-7 C