Question

What is the force between two small charged spheres having
charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?

Answer 1

$Q_1=2 \times 10^{-7}C\\Q_2=3 \times 10^{-7}C\\R=30cm=0.3m\\ \\F= \frac{1}{4 \pi \epsilon_o}\ \frac{Q_1Q_2}{R^2}=9 \times 10^9 \times \frac{2 \times 10^{-7} \times 3 \times 10^{-7}}{0.3^2} =\boxed{0.006N}$

Force acting between the charges is 0.006N and it is repulsive(since it is positive)

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

q1 = 2 × 10^−7 C

q2 = 3 × 10^−7 C

r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

F= (q1 q2)/(4πε°r^2)

Where, ε° = Permittivity of free space

1/4πε° = 9× 10^9 Nm^2C^-2

So,

F= (9×10^9 × 2× 10^-7 × 3× 10^-7)/(0.3)^2

= 6× 10^-3 N

Hence,

⭐force between the two small charged spheres is 6 × 10^−3 N

⭐Both charge have same nature. Hence, force between them will be repulsive

I hope, this will help you.

Thank you______❤

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