Question

A small charged bead has a mass of 3.4 g . It is held in a uniform electric field E⃗ = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 14 m/s2 .

What is the charge on the bead?

Answer 1

Solution -:

Then applied electric field is F = qE

So , net F = ma

So , qE - mg = ma

qE = mg+ma

q = m(g+a)/E

Here , m = 3.4 g = 3.4*10^-3 kg

g = 9.8 m/s^2

Acceleration, a = 14 m/s^2

E = 200000 N/C

Put the values..

q = ((3.4 * 10^-3) * (9.8+14)) / 200000

q = 4.046*10^-7 C (Answer)

I hope help you !!