A box is placed on the floor. The area of the box in contact with the floor is 2.4m²Pressure exerted on the floor 16 newtons/m²Work out the force exerted by the box on the floor

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Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then   

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge

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