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AURORKA [14]
4 years ago
15

A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no

w removed. After sphere A has been removed, sphere B is touched to an initially uncharged sphere, sphere C. What is the sign of the charge, if any, on sphere C after it has been touched by sphere B
Physics
1 answer:
Yuri [45]4 years ago
3 0

Answer:

The sphere C carries no net charge.

Explanation:

  • When brougth close to the charged sphere A, as charges can move freely in  a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
  • If in such state, if  the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of  charge principle, no net  charge can be built on sphere C.
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v=35m/s

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How long must it take for a communications satellite to make one complete orbit around the earth? (such an orbit is said to be g
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2 years ago
i will give the brainliest answer to whoever answers this, explain how Sir William Gilbert used models in his investigation of m
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7 0
3 years ago
The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.
oksano4ka [1.4K]

Answer:

i = 2.84 \times 10^{-3} A

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

J = \frac{di}{dA}

so the current through the wire is given as

i = \int J dA

now we have

i = \int_{0.921R}^R J dA

here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr

now we have

i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr

i = (2.04 \times 10^9) \frac{r^4}{4}

now plug in both limits as mentioned

i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})

i = (2.04\times 10^9)(0.07 R^4)

here R = 2.11 mm

i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)

i = 2.84 \times 10^{-3} A

8 0
3 years ago
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