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Ch 22: Gauss' Law

Chapter 22, Problem 16

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate: (b) the electric field at the planet's surface (refer to the astronomical data inside the back cover);

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Welcome back everybody. We are observing a spherical planet here and we are told that it has a radius of 3200 kilometers or 3200 times 10 to the third meters. We are also told that it generates an electric flux at the surface of negative 2.14 times 10 to the 18th. Newton meters squared per cool. Um, and we are tasked with finding what the magnitude and direction is of the electric field near the surface of the planet. Well, we know that the electric flux or a uniform electric field is equal to the magnitude of the electric field times the area times the cosine of the angle between the flux and the quote unquote normal vector of the surface here. So say we have our planet and it kind of looks like this. This will be our normal vector. Now we can go ahead and determine the direction. Right now, we see that our flux is negative, meaning that the flux is going to be acting towards the center of the planet, meaning that the angle measure between these two vectors is a full 180°. Now that we've established that I want to go back to our equation up here, we need this e term. So I'm gonna go ahead and isolate that real quick. I'm gonna divide both sides by the area, times the cosine of our omega area, Kassian omega and all of this will cancel out on the right hand side, leaving us with that the electric field is equal to the electric flux divided by the area, the surface area times the cosine of our angle. We have all of these values, so let's go ahead and plug them in. We have that are electric flux is negative. 2.14 times 10 to the 18th. This is all divided by the surface area of our spherical planet. So the surface area of a sphere is four pi times the radius squared radius being 3200 times 10 to the third meters. Weird. And then this is multiplied by our co sign of our angle, which we established to be 180 degrees. This of course is going to be equal to 1. times 10 to the fourth. Newtons per cool. Um and as we said, since the flux is negative, it is directed towards the center of the planet. Giving us our final answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.
Related Practice
Textbook Question
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?
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Textbook Question
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. (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?
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Textbook Question
Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate: (a) the total electric charge on the planet;
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Textbook Question
Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate:(c) the charge density on Mars, assuming all the charge is uniformly distributed over the planet's surface.
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Textbook Question
A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere?
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Textbook Question
A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (b) Calculate the strength of the electric field just outside the sphere?
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