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tino4ka555 [31]
2 years ago
6

The light given off by an object based only on its temperature supports the

Physics
2 answers:
liq [111]2 years ago
5 0

Answer:

A.Blackbody Radiation

Explanation:

UkoKoshka [18]2 years ago
5 0

Answer:

A. Blackbody radiation

Explanation:

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A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the
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(b). The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Explanation:

Given that,

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(a). We need to calculate the potential at a distance r = 1.60 cm

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V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr

V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}

V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

V_{r}=-0.00012190\ V

V_{r}=-1.219\times10^{-4}\ V

The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

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V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}

V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

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