A-3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m.

a. find the electric field (magnitude and direction) at point on the x-axis at x = 0.200 m.

express your answer with the appropriate units. enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.

b. find the electric field (magnitude and direction) at point on the x-axis at x = 1.20 m.

express your answer with the appropriate units. enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.

c. find the electric field (magnitude and direction) at point on the x-axis at x = -0.200 m.

express your answer with the appropriate units. enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.

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

So are u adding some additional catalyst