A hollow, conducting sphere with an outer radius of 0.250 m and a...

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?

Accepted Answer

Hey, everyone. So this problem is dealing with electric fields. Let's see what it's asking us. We have a point charge of negative five micro columns held at the center of a thin hollow spherical shell. The shell has an internal radius of seven centimeters, an external radius of eight centimeters and an initial surface charge density of 20 nano columns per centimeter squared. We're asked to find the magnitude of the electrical field near the surface of the shell. Our answers in units of newtons per Coolum R A 1.54 times 10 to the three B 1.93 times 10 to the four C 1. times 10 to the seven or D 1.93 times 10 to the nine. So we can recall that our flux is given by five E equals Q enclosed divided by epsilon knot. And that is equal to E A. So when we're solving for E, we can isolate that variable and it becomes Q enclosed divided by epsilon null epsilon knot. Our electric constant multiplied by A R area are Q enclosed are in charge, are enclosed. Charge is going to equal the initial charge minus this point charge that was added to it. So minus five micro colons, we can recall that the equation for Q is given by sigma multiplied by area where sigma is the surface charge density. So solving for our initial QE we have 20 times 10 to the negative nine cool arms and then to go from centimeters squared to meters squared, it'll be 10 to the minus four or sorry 10 to the negative m squared. Multiplying that by our area, we can recall the area of a circle is four pi R squared. So four pi and then we're looking at the boundary of the sphere which was given to us as eight centimeters. And so keeping in standard units, I'm writing that as 0.08 m and that quantity squared, can we plug that in? Can we get 1.61 times 10 to the negative five coons? So now we can solve for our Q enclosed our enclosed charge 1.61 times 10 to the negative five columns minus five times 10 to the negative six columns. A micro column um is 10 to the negative six and we get 1.11 times 10 to the negative five polus. And so now we can finally solve for our electric field magnitude E equals it's Q and close 1.11 times 10 to the negative five polus divided by epsilon, not our electric constant. We can recall is 8. oops 8. times 10 to the negative 12 Pullum squared per newton meter squared, multiplied by area again, for pi multiplied by 0. m squared. We plug that in to our calculators and we get an electric field magnitude of 1.54 times 10 to the seven newtons per colum. And when we look at our multiple choice answers that aligns with answer choice C so C is the correct answer for this problem. That's all we have for this one, we'll see you in the next video.

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