The magnitude of an electric field is 100 newtons per coulomb at a distance of 0.50 meters from a point particle that has a charge of 𝑞. What is the value of 𝑞?
We’re told the electric field magnitude is 100 newtons per coulomb, which we’ll call capital 𝐸, and that that is the value of the electric field a distance of 0.50 meters from a point particle 𝑞. We’ll call that distance 𝑑.
We want to solve for the value of 𝑞. As a start, we can recall the relationship for an electric field caused by a point particle. The electric field that created by a point particle with charge 𝑞 is equal to that charge times Coulomb’s constant, 𝑘, divided by the distance squared between the charge, 𝑞, and where the electric field, 𝐸, is being experienced or measured.
In this exercise, we’ll treat 𝑘 as exactly 8.99 times 10 to the ninth newton meter squared per coulomb squared. When we write the equation for electric field in terms of the variables in our situation, we see we can rearrange this equation to solve for 𝑞.
𝑞 equals the electric field times the distance squared divided by 𝑘, the constant. When we plug in for 𝐸, 𝑑, and 𝑘 and enter these values on our calculator, we find that 𝑞, the charge creating this electric field, has a value of 2.8 times 10 to the negative ninth coulombs. That is the value of the point charge 𝑞.