a hollow non conducting spherical shell has inner radius r1 7 cm and outer radius r2 13 cm a cha

Step 1: We need to find the total charge on the shell. The charge density ρ is given by A*r, so

A hollow non conducting spherical shell has inner radius r1...

A hollow non-conducting spherical shell has an inner radius R1 = 7 cm and an outer radius R2 = 13 cm. A charge Q = -35 nC lies at the center of the shell. The shell carries a spherically symmetric charge density ρ = Ar for R1 < r < R2 that increases linearly with radius, where A = 18 μC/m^4. Part D: What is the radial electric field at the point r = 2R2? Give your answer in units of kN/C.

Transcript

00:01 Hi, students in this question, a hollow non -conducting spherical shell as shown in the figure is been given, and we are asked to find out the radial electric field e at a distance r2 is twice the value of outer radius r2.

00:13 And in this figure, you can see that the outer radius is taken as r2 and the inner radius is taken as r1.

00:20 And the value of inner radius r1 is taken to be 7 cm.

00:24 On the other hand, the value of outer radius r2 is taken to be 13 cm.

00:30 Now here, since we have to find out the value of electric field for the case of small letter r to be equal to twice the value of outer radius r2, we can find out the value of r to be equal to twice the value of 13 which will be equal to 26 cm.

00:44 Now in order to find the value of electric field, we have to use the gauze slope which is given as surface integral of electric field dot da where da corresponds the area will be equal to 1 by epsilon 0 into total charge q enclosed inside that particular shell.

01:03 And here in this question, the value of this charge q is taken to be minus 35 into 10 raised to minus 9 kolo.

01:12 Now in this question, since we have to find out the value of radial electric field, our go so slow changes as e .da to be equal to 1 by epsilon 0 into q plus integral of r1 to r2, row dot da into dr, where this this da corresponds to the area of the spear which is taken as 4 pi r square.

01:32 On the other hand, row corresponds to the charge density which is equal to a product of constant a into distance from the center small letter r.

01:40 And here by substituting the value of da in the above equation, our equation changes as e into 4 pi r square will be equal to 1 by epsilon 0 into cube plus integral of r1 to r2...

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