Respuesta :
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
[tex]F_e = \dfrac{kQ_1Q_2}{r^2}[/tex]
[tex]F_e = \dfrac{9\times 10^9\times (-21.0 \times 10^{-9})^2}{0.1^2}[/tex]
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
[tex]F_{net} = m g - F_e[/tex]
[tex]F_{net} = 0.19 \times 10^{-3}\times 9.8 - 3.96\times 10^{-4}[/tex]
[tex]F_{net} = 1.466\times 10^{-3}\ N[/tex]
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:
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
[tex]F=\dfrac{kQ^2}{d^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{9\times10^{9}\times(-22\times10^{-9})^2}{(10\times10^{-2})^2}[/tex]
[tex]F=4.356\times10^{-4}\ N[/tex]
We need to calculate the net force
Using formula of net force
[tex]F=weight- F_{charge}[/tex]
Put the value into the formula
[tex]F=0.19\times10^{-3}\times9.8-4.356\times10^{-4}[/tex]
[tex]F=0.0014264\ N[/tex]
[tex]F=1.4264\times10^{-3}\ N[/tex]
We need to calculate the magnitude of its initial acceleration
Using newton's second law of motion
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.4264\times10^{-3}}{0.19\times10^{-3}}[/tex]
[tex]a=7.50\ m/s^2[/tex]
Hence, The magnitude of its initial acceleration is 7.50 m/s²
