Question

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at What is the magnitude of the charge on each sphere, assuming only that the electric force is present? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

Answer 1

Answer:

$1.36\times 10^{-7} C$

Explanation:

We are given that

Mass of charged tine spheres=m=1 g=$\frac{1}{1000}=0.001 kg$

1 kg=1000g

The distance between charged tine spheres=r=2 cm=$\frac{2}{100}=0.02 m$

1 m=100 cm

Acceleration =$a =414 m/s^2$

Let q be the charge on each sphere.

$k=9\times 10^9Nm^2/C^2$

The electric force between two charged particle

$F=\frac{kq_1q_2}{r^2}$

Using the formula

The force between two charged tiny spheres=$F_e=\frac{kq^2}{(0.02)^2}$

According to  Newton's second law , the net force

$F=ma$

$F=F_e$

$0.001\times 414=\frac{9\times 10^9\times q^2}{(0.02)^2}$

$q^2=\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}$

$q=\sqrt{\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}}$

$q=1.36\times 10^{-7} C$

Hence, the magnitude of charge on each tiny sphere=$1.36\times 10^{-7} C$