# Video: Relating a Uniform Electric Field to the Energy of a Charge in the Field

An ion with twice the charge of an electron is accelerated from rest to a kinetic energy of 32.0 keV by the electric field between two parallel conducting plates. If the plates are separated by a distance of 2.00 cm, what is the electric field strength between the plates? Proton charge is +1.60 × 10⁻¹⁹ C.

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### Video Transcript

An ion with twice the charge of an electron is accelerated from rest to a kinetic energy of 32.0 kiloelectron volts by the electric field between two parallel conducting plates. If the plates are separated by a distance of 2.00 centimetres, what is the electric field strength between the plates? Proton charge is plus 1.60 times 10 to the negative 19th coulombs.

We can call the energy of the accelerated electron 32.0 kiloelectron volts capital 𝑈. And we can call the distance between the parallel plates 2.00 centimetres 𝑑. We want to figure out the electric field strength between the plates. We’ll call that 𝐸.

We can start off by drawing a diagram of this scenario. We have two oppositely charged parallel plates separated by a distance 𝑑. And the opposite charges on the plates create a uniform electric field 𝐸 between the two. Our point charge which we’re told is somewhere in between the plates begins at rest and then due to the electric field is accelerated and gains an energy of 32.0 kiloelectron volts. If we call our charge 𝑄, we’re told that 𝑄 is twice the charge of an electron.

Based on this information, we want to solve for the magnitude of the electric field 𝐸 in between the parallel plates. To begin, we can recall two mathematical relationships. The first has to do with electric potential 𝑉. In the case of our parallel conductor, the electric potential between the plates is equal to the electric field between the plates times the distance separating them.

The second relationship that we recall has to do with electric potential energy. This term is equal to electric potential multiplied by the charge involved. In our case, we can combine these two equations by inserting 𝑉 equals 𝐸𝑑 in for 𝑉 in our equation for electric potential energy. If we rearrange this equation to solve for 𝐸, we see it equals electric potential energy divided by charge times the distance between the plates.

In this equation, the charge lowercase 𝑞 represents the charge capital 𝑄 in our problem. So we can make that substitution and then notice that we have both 𝑈, 𝑄, and 𝑑 given to us in the problem statement. So we’re ready to plug in and solve for 𝐸.

Before entering these values on our calculator, let’s notice a few things about these values. First, we’ve used units of electron volts in our numerator and units of metres in our distance in the denominator. Speaking of electron volts, an electron volt is defined as the energy gained by the charge of one single electron, moving across the potential difference of one volt. We’re told that the charge of our ion is two times the charge of an electron.

Rather than writing this charge into our denominator and including it in our calculation, we in fact we can cancel it out because of our electron volts numerator. So we transition from electron volts to volts eliminating the charge of the electron in the denominator. We’re now ready to calculate 𝐸. And when we do, we find that to three significant figures it’s 8.00 times 10 to the fifth volts per metre. That’s the electric field that exists between the two parallel plates.