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3 years ago

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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 o

ther 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)

1 answer:

KATRIN_1 [288]3 years ago

4 0

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

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