A capacitor is formed from two concentric spherical...

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 10.5 cmcm , and the outer sphere has radius 16.5 cmcm . A potential difference of 140 VV is applied to the capacitor. Part A What is the energy density at rr= 10.6 cmcm , just outside the inner sphere? Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type uu = 3.78•10−5 J/m3J/m3

Transcript

00:01 In the given problem we have been given a spherical capacitor which comprises of two spherical shells as shown here one outer and one inner here the radii of these two shells are given as for inner one it is 12 .5 centimeter or we can see 12 .5 into 10 dash per minus 2 meter for the outer one here this is inner and this is outer for outer one this is 14 .8 centimeter or 14 .8 into 10 dash per minus 2 meter first of all we will find the capacitance of such a capacitor using the expression 4 pi epsilon not a b upon a minus b upon b minus a and when i put these values the capacitance of such a capacitor will come out to be 89 .4 pico ferret now the potential across the plates across the shells has been given as 120 volt so now we will find charge stored over such a capacitor which is given as c into v so it will be 89 .4 into 10 dashed bar minus 12 ferret into 120 volt so the charge is told comes out to be 1 .074 into 10 dash bar minus 8 coularm now we have to find electric fields at two points first of the point is very close to the inertial and distance r1 and second point is very close to the outer shell at a distance r2 and these distances are given as r1 is 12 .6 centimeter and r2 is 14 .7 centimeter we have to find energy densities at these two points so to find energy density first of all we will find electric field intensities at these two points to find electric field intensity we will use the expression 1 upon 4 pi epsilon not q by r square so for even it will become 9 into 10 ratio per 9 into q 1 .074 into 10 dash to the power minus 8 divided by 12 .6 into 10 dash to bar minus 2 to the whole square and this electric field comes out to be 6 ,088 volt per meter similarly we will find electric field at the second point nine into ten days to the power nine 1 .074 into 10 to the power minus 8 and divided by 14 .7 into 10 dash to the part minus two the whole square and this electric field comes out to be 4 ,473 volt per meter so now at the first part of the problem we can find energy density at first point new one which comes out to be half epsilon not e square is its standard expression so here half into 8 .854 into 10 dash minus 12 into e square means this is 6088 to the whole square and this energy density comes out to be 1 .64 into 10 dash to the power minus 4 joules per meter cube in the second part of the problem we have to find energy density at the second point so it will be given as half epsilon node e2 square so it is half into 8 .854 into 10 dash to power minus 12 4, 4 ,000 4 ,000 173 to the whole square and it comes out to be 8 .85 into 10 dash of minus 5 joules per meter cube...

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