Answer:
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
Electric field due to plates
Ef = V/d
Ef = 2033 V / (2.0 * 10^-2 m )
Ef = 101650 V/m
So, we can write
Ef * q = m*g
q = m*g / E
f
The mass can be equal using the density and the volume so:
m = ρ * v
The volume can be find as:
v = 2.298 x 10 ⁻ ¹⁶ m³
q = ρ * v * g / Ef
q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
By Hooke's Law,
F=kx
The only force acting here is weight, and x is the extension of the string (you need to convert this to mm) so
mg=kx
(0.15)(9.81)=k ((420-300)x10^-3)
Then just solve this equation.
