# Mass of charged sphere suspended between charged plates

## Homework Statement

A small sphere with charge 2.4 micro coulombs is suspended from a thread between 2 charged plates. The plates have a voltage of 62 and the distance between the plates is 3.1 cm. The sphere hangs at 18 degrees to the vertical.

E = V/r
FE = Eq
Fnetx = FE - FT
Fnety = FTy - Fg

## The Attempt at a Solution

[/B]
E = 62V/0.031m = 2000 N/C

FE = Eq = (2000 N/C)(2.3 x 10^-6 C)

FTx = FE = 4.6 x 10^-3 N

FTy = FE * tan18 degrees
= 4.6 x 10^-3 N * tan18
= 1.5 x 10^-3 N

Fg = mg
1.5 x 10^-3 N = m(9.8 m/s^2)
m = 1.5 x 10^-4 kg

TSny
Homework Helper
Gold Member
Hello. In the problem you state that the charge is 2.4 μC but you use 2.3 μC in the calculation.

You wrote FTy = FE * tan18o . Is this correct (keeping in mind that the 18o is measured from the vertical)?

Hello. In the problem you state that the charge is 2.4 μC but you use 2.3 μC in the calculation.

You wrote FTy = FE * tan18o . Is this correct (keeping in mind that the 18o is measured from the vertical)?

I just realised the problem statement says 2.4 μC but the diagram says 2.3 μC hmmm. Assuming 2.4 is correct..

FTy = FE*tan72
= 1.5 * 10^-2 N

Fg = mg
1.5 * 10^-2 = m(9.8)
m = 1.5 ^ 10^-3 kg

TSny
Homework Helper
Gold Member
That looks right.

That looks right.

Ok than you!