Question

A small metal sphere has a mass of 0.14 g and a charge of -26.0 nC. It is 10.0 cm directly above an identical sphere that has the same charge. This lower sphere is fixed and cannot move. If the upper sphere is released, it will begin to fall.Part AWhat is the magnitude of the force between the spheres?Part BIf the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?

Answer 1

Answer:

A)

$608.4\times10^{-6} N$

B)

$5.5 ms^{-2}$

Explanation:

A)

$q$ = magnitude of charge on each sphere = $26\times10^{-9} C$

$r$ = Distance between the two spheres = 10 cm = 0.10 m

$F$ = magnitude of force between the two spheres

Using Coulomb's law, magnitude of the force between two charged sphere

$F = \frac{kq^{2}}{r^{2}}\\F = \frac{(9\times10^{9})(26\times10^{-9})^{2}}{(0.1)^{2}}\\F = 608.4\times10^{-6} N$

B)

$m$ = mass of the sphere = $0.14\times10^{-3} kg$

$F_{g}$ = Force of gravity in down direction = $mg = (0.14\times10^{-3})(9.8) = 1.372\times10^{-3} N$

$F$ = Electrostatic force of repulsion in upward direction = $608.4\times10^{-6} N$

$a$ = acceleration of the sphere

Force equation for the motion of the sphere is given as

$F_{g} - F = ma \\1.372\times10^{-3} - 608.4\times10^{-6} = (0.14\times10^{-3}) a\\a = 5.5 ms^{-2}$

Answer 2

Answer:

(a) 6.084 x 10^-13 N

(b) 97.87 m/s^2

Explanation:

mass, m = 0.14 g

charge, q = - 26 nC = 26 x 10^-19 C

d = 10 cm = 0.1 m

(a) the force between the two is electrostatic force.

$F_{e}=\frac{Kq^{2}}{d^{2}}$

$F_{e}=\frac{9\times 10^{9}\times 26\times 26\times 10^{-18}}{0.01}$

Fe = 6.084 x 10^-13 N

(b) the gravitational force of mass

Fg = m x g = 0.14 x 10^-3 x 9.8

Fg = 137.2 x 10^-5 N

Net force acting on the mass

Fg - Fe = ma

137.2 x 10^-5 - 6.084 x 10^-13 = 0.14 x 10^-3 x a

a = 97.87 m/s^2