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A metal sphere with radius ra is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb. There is charge +q on the inner sphere and charge −q on the outer spherical shell. Take V to be zero when r is infinite.


a. Calculate the potential V(r) for r
b. Calculate the potential V(r) for ra
c. Calculate the potential V(r) for r>rb

d. Find the potential of the inner sphere with respect to the outer.


Use the equation Er=−∂V∂r and the result from part (b) to find the electric field at any point between the spheres (ra

Use the equation Er=−∂V∂r and the result from part (c) to find the electric field at a point outside the larger sphere at a distance r from the center, where r>

Respuesta :

Answer:

a. kq (1/ra-1/rb)

b. Kq (1/r-1/rb)

c. 0

d. Kq(1/ra-1/rb)

Explanation:

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(a) The potential for r is, V(r) = kq/r.

(b) The potential for [tex]r_{a}[/tex] is [tex]V(r= r_{a})=\dfrac{kq}{r_{a}}[/tex].

(c)  The potential for [tex]r>r_{b}[/tex] is, [tex]V(r>r_{b})=\dfrac{-kq}{r_{b}}[/tex] .

(d)  The potential of the inner sphere with respect to the outer is [tex]kq ( \dfrac{1}{r_{a}} -\dfrac{1}{r_{b}})[/tex].

Given data:

The radius of sphere is, [tex]r_{a}[/tex].

The radius of spherical shell is, [tex]r_{b}[/tex].

The magnitude of charge at inner and outer is, +q and -q.

(a)

Since, inner sphere is positively charge. So, the expression for the electric potential V(r) for r, due to inner metallic sphere is as follows:

[tex]V(r)=\dfrac{kq}{r}[/tex]

Here, k is the Coulomb's constant.

Thus, the potential for r is, V(r) = kq/r.

(b)

The potential for radius of sphere [tex](r_{a})[/tex] is obtained by substituting [tex]r = r_{a}[/tex].

So, the required potential is,

[tex]V(r= r_{a})=\dfrac{kq}{r_{a}}[/tex]

Thus, the potential for [tex]r_{a}[/tex] is [tex]V(r= r_{a})=\dfrac{kq}{r_{a}}[/tex].

(c)

For the potential corresponding to [tex]r>r_{b}[/tex], the electric potential is given as,

[tex]V(r>r_{b})=\dfrac{k(-q)}{r_{b}}\\\\V(r>r_{b})=\dfrac{-kq}{r_{b}}[/tex]

the negative sign shows that the charge at outer sphere has negative polarity.

Thus, the potential for [tex]r>r_{b}[/tex] is, [tex]V(r>r_{b})=\dfrac{-kq}{r_{b}}[/tex] .

(d)

Now, the potential of the inner sphere with respect to the outer is given as,

[tex]V'= V(r=r_{a})+V(r>r_{b})\\\\V'= \dfrac{kq}{r_{a}} +\dfrac{k(-q)}{r_{b}}\\\\V'= \dfrac{kq}{r_{a}} -\dfrac{kq}{r_{b}}\\\\V'=kq ( \dfrac{1}{r_{a}} -\dfrac{1}{r_{b}})[/tex]

Thus, the potential of the inner sphere with respect to the outer is [tex]kq ( \dfrac{1}{r_{a}} -\dfrac{1}{r_{b}})[/tex].

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