Two charges +2 coulomb each are placed 2 m apart in vacuum, force

SOLUTION

CONCEPT:

• Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 109 Nm2/C2

CALCULATION:

Given that:

q₁ = 2 C and q₂ = 2 C, r = 2 m

The force exerted by one charge q₁ on another charge q₂ is given by Coulomb's law:

\(F = \frac{1}{{4\pi {_0}}}\frac{{q_1 q_2}}{{{r^2}}}\;\)where \(\frac{1}{{4\pi {_0}}} = 9 \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$\)

\(F = 9 \times {10^9} \times \frac{{2 \times 2}}{{{2^2}}} = 9 \times {10^9}N\)

So option 1 is correct.