Question

A small metal sphere has a mass of 0.19 g and a charge of -21.0 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.
What is the magnitude of its initial acceleration?

Answer 1

Answer:

a =  7.71 m/s²

Explanation:

given,

mass of the small sphere = 0.19 g

charges Q₁ and Q₂ = -21.0 n C

distance between the charge, r = 10 cm

                                                     = 0.1 m

magnitude of acceleration = ?

Calculation of force between the two charges

$F_e = \dfrac{kQ_1Q_2}{r^2}$

$F_e = \dfrac{9\times 10^9\times (-21.0 \times 10^{-9})^2}{0.1^2}$

    F_e = 3.96 x 10⁻⁴ N

now,

On releasing there will be two force acting, one is due to weight of the sphere and other is the electric force.

Net force on sphere = Weight - Fe

$F_{net} = m g - F_e$

$F_{net} = 0.19 \times 10^{-3}\times 9.8 - 3.96\times 10^{-4}$

$F_{net} = 1.466\times 10^{-3}\ N$

now,

F = m a

1.466 x 10⁻³ = 0.19 x 10⁻³  x a

a =  7.71 m/s²

the magnitude of its initial acceleration is given by 7.71 m/s²

Answer 2

Answer:

The magnitude of its initial acceleration is 7.50 m/s²

Explanation:

Given that,

Mass of sphere = 0.19 g

Charge = -21.0 nC

Distance = 10.0 cm

We need to calculate the electric force

Using formula of force

$F=\dfrac{kQ^2}{d^2}$

Put the value into the formula

$F=\dfrac{9\times10^{9}\times(-22\times10^{-9})^2}{(10\times10^{-2})^2}$

$F=4.356\times10^{-4}\ N$

We need to calculate the net force

Using formula of net force

$F=weight- F_{charge}$

Put the value into the formula

$F=0.19\times10^{-3}\times9.8-4.356\times10^{-4}$

$F=0.0014264\ N$

$F=1.4264\times10^{-3}\ N$

We need to calculate the magnitude of its initial acceleration

Using newton's second law of motion

$F=ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.4264\times10^{-3}}{0.19\times10^{-3}}$

$a=7.50\ m/s^2$

Hence, The magnitude of its initial acceleration is 7.50 m/s²