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klemol [59]
2 years ago
5

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti

al is 150 V. What is the radius of the sphere
Physics
1 answer:
igomit [66]2 years ago
4 0

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, V_s = 450 V

potential at radial distance, V_r = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}

450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r}  \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m

Therefore, the radius of the sphere is 4.05 m

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Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

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Kinetic energy is given as;

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K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

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3 years ago
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