The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4×10−15 m. (a) What is the electric field this nucleus produces just outside its surface?

Video transcript

Welcome back everybody. We are taking a look at the Earth and we are told that it is approximately a sphere with a radius of 6.37 times 10 to the six m. Now, together with the atmosphere, it is electrically neutral. We are told that the earth carries a negative charge. However, so Q. Of E. Of negative 6.8 times 10 to the fifth column. And we are tasked with finding what the magnitude of the electric field is produced by the Earth at a location near its surface. So, using God's Law here, we know that the electric flux is equal to the electric field, vector times the surface area. Now this is the same thing as negative E since the electric flux in this case is going to be directed into the Gaussian surface times four pi R squared, which is just the surface area of the sphere. We also know from God's law that the flux is equal to the charge of the earth divided by the electric utility constant. So we're going to combine these two equations into one equation and we get that E times for, sorry, negative E times for pi R squared is equal to the queue of the Earth. Charge of the Earth divided by the electric purgative itty constant. Now, I am going to divide both sides by negative four pi r squared negative four I R squared. All of these terms will cancel out and we get that are electric field is equal to this big fraction right here, I'm actually gonna separate this into two fractions and you'll see why a little bit later on. So I'm gonna separate this into negative 1/4 pi times r epsilon not times the charge of the earth, All divided by the radius squared. Now that we have our formula for the magnitude of our electric field, let's go ahead and plug in some numbers. So this is gonna be negative whatever this term is. Well, this term 1/4 pi epsilon is actually equivalent to coolants constant. So this is going to be 8.99 times 10 to the ninth. Now this will be times negative 6.8 times 10 to the fifth, which is the charge of Earth divided by the radius of earth, which is 6.37 times 10 to the six. All squared. When you plug this into your calculator, you get that the magnitude of the electric field produced by Earth is 150.6 newtons per Coolum corresponding to answer choice. C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one