A metal sphere with radius ra is supported on an insulating...

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ . There is charge $+$q on the inner sphere and charge $-$q on the outer spherical shell. (a) Calculate the potential $V(r)$ for (i) $r < r_a ;$ (ii) $r_a < r < r_b ;$ (iii) $r > r_b .$ ($Hint$: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{ab} = \frac{q} {4\pi\epsilon_0} ( \frac{1} {r_a} - \frac{1} {r_b} )$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r) = \frac {V_{ab}} {(1/r_a - 1/r_b)}\frac {1} {r_2}$$ (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r > r_b$ . (e) Suppose the charge on the outer sphere is not $-q$ but a negative charge of different magnitude, say $-Q$. Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

Transcript

00:01 We know that the electric potential outside the charged sphere is v equal to 1 upon 4 pi epsilon or we can say 1 divided by 4 pi epsilon node, 2 divided by 4 pi epsilon node, 2 divided by r.

00:44 Now here q is the charge on the metal sphere and r is the distance from the center of the sphere to the fade point.

00:57 And 1 divided by 4 pi epsilon node is represent the coulomb's constraint.

01:03 Now similarly, the electric potential, potential inside the charged metal sphere is given by v equal to 1 divided by 4 pi epsilon 0 q divided by capital r now comes to part first.

01:49 So here we are going to calculate the total electric potential in the reason where r is less than r a.

02:04 So for reason, r less than ra, so for reason r less than ra, that that is inside the metal sphere, the electric potential due to metal sphere is v1 of r equal to k, q divided by r .a.

02:44 Now here, r .a is the radius of the matter sphere charged on the outer holocair.

02:53 And similarly, for reason, are less than r a that is inside the hollow sphere the electric potential due to hollow sphere is given by v2 of r equal to minus k q divided by r b now here r b is the radius of the polo sphere thus the total electric potential is equal to the sum of the electric potential due to metal sphere and electric potential due to hollow sphere that is v of r is equal to v1 of r plus v2 of r now on substitute the values we get k q divided by r a minus k q divided by r b b b by r b b b now we can write this as k q multiply by 1 divided by r a minus 1 divided by r b now we know that k is equal to 1 divided by 4 pi epsilon naught so we can write the above relation as v of r is equal to q divided by 4 pi epsilon not multiply by 1 divided by r a minus 1 divided by r b.

05:33 Therefore, the total electric potential in the reason r less than r a is q divided by 4 pi epsilon multiply by 1 divided by r a minus 1 divided by r b.

05:53 Now comes to part second.

05:56 So here we are going to calculate the total electric potential in the reason, where r a is less than r and r is less than r b so first for reason r a less than r and r less than r b that is in the outside the metal sphere the electric potential due to metal sphere is b1 of r equal to k q divided by r.

06:50 Now for reason r a less than r less than r b, that is inside the hollow.

07:12 Sphere the electric potential due to holosphere can be written as b2 of r equal to minus k q divided by r b now here r b is the radius of the hollow sphere thus the total electric potential is equal to the sum of the of the electric potential due to metal stair and electric potential due to holo sphere.

07:50 That is, v of r is equal to v1 of r plus v2 of r.

08:01 Now, and substitute the values we get k -k divided by r minus k -k cubed divided by r now we can write this as kq multiply by 1 divided by r minus 1 divided by r b.

08:30 Now since k is equal to 1 divided by 4 pi epsilon not, therefore v.

08:43 R is equal to q divided by 4 pi epsilon note multiply by 1 divided by r minus 1 divided by r b b.

09:01 Therefore the total electric potential in the reason r a less than r r r is q divided by 4 pi epsilon by 1 divided by r minus 1 divided by r b now comes to our third so here for reason are greater than r b that is outside the metal sphere the the electric potential due to metal spheres can be written as v1 of r equal to k q divided by r...

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