The electric field strength in a particular thundercloud is 2.0 times 10 to the fifth newtons per coulomb. What is the magnitude of the acceleration of an electron in this field?
We can call the electric field strength, 2.0 times 10 to the fifth newtons per coulomb, capital 𝐸. We want to know how much an electron would accelerate in this field considering only the magnitude of that acceleration. We’ll call this magnitude of lower case 𝑎. To start our solution, let’s recall Newton’s second law.
This law says that the net force acting on an object is equal to that object’s mass times its acceleration. If we call the force acting on the electron 𝐹 sub 𝑒, then that force by the second law is equal to the electrons mass, 𝑚 sub 𝑒, times its acceleration, 𝑎.
Now because the force in this example is an electric force, we can also recall from Coulomb’s law that the electric force between two point charges 𝑞 one and 𝑞 two equals their product times 𝑘, the constant, divided by the distance between them squared.
If we take one of these two charges, say we take 𝑞 two, and considerate the electric field that 𝑞 two generates and call that 𝐸, then the force is also equal to 𝑞 one times 𝐸. In other words, using these particular symbols, the electric field 𝐸 equals 𝑘 times 𝑞 two divided by 𝑟 squared. We can use this alternative formulation of Coulomb’s law in our exercise.
The force, 𝐹 sub 𝑒, on the electron equals its mass times its acceleration. And it also equals the charge of the electron, 𝑞 sub 𝑒, times the magnitude of the electric field that it’s in. Regarding the electron, we treat its mass as exactly 9.1 times 10 to the negative 31st kilograms. And we consider its charge exactly negative 1.6 times 10 to the negative 19th coulombs.
If we take the equation that results from combining Newton’s second law with Coulomb’s law and rearrange to solve for acceleration, 𝑎, we find that it’s equal to the electron charge divided by the electron mass times the electric field 𝐸. Since we want to solve not just for the electrons acceleration but the magnitude of its acceleration, let’s put absolute value bars around both sides of the equation.
Since we have values for 𝑞 sub 𝑒, 𝑚 sub 𝑒, and 𝐸, we’re ready to plug in and solve for the magnitude of 𝑎. With these values entered in, we can now put them in our calculator and solve for 𝑎. We find that, to two significant figures, 𝑎 is 3.5 times 10 to 16th meters per second squared. That’s the magnitude of the acceleration experienced by the electron in this electric field.