The strength of the earth’s magnetic field b at the equator is approximately equal to 5 × 10−5 t. the force on a charge q moving in a direction perpendicular to a magnetic field is given by f = q v b, where v is the speed of the particle. the direction of the force is given by the right-hand rule. suppose you rub a balloon in your hair and your head acquires a static charge of 9 × 10−9 c. if you are at the equator and driving west at a speed of 90 m/s, what is the strength of the magnetic force on your head due to the earth’s magnetic field? answer in units of n. what is the direction of that magnetic force?

Aliquid is an answer i hope this

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