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...
Question 1 Report
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 Newton and couloumb
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.
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