A charge of 1 x \(10^{-5}\)C experiences a force of 40N at a certain point in space. What is the magnitude of the electric field intensity at the point in N...

Answer Details

The electric field intensity at a point is defined as the electric force per unit charge experienced by a charge placed at that point. Given, a charge of 1 x \(10^{-5}\)C experiences a force of 40N at a certain point in space. So, electric field intensity (E) at that point can be calculated as follows: E = F/q where F is the force experienced by the charge q. Substituting the given values, we get: E = 40 N / (1 x \(10^{-5}\)C) E = 4 x \(10^{6}\) N/C Therefore, the magnitude of the electric field intensity at the point is 4 x \(10^{6}\) N/C. Note: Coulomb is the unit of charge, not electric field intensity.