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slamgirl [31]
3 years ago
5

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

press your answer in terms of the given quantities and appropriate constants.1. What is the electric field at the surface of the sphere?E=2.What is the electric potential at the surface of the sphere?V=
Physics
1 answer:
amid [387]3 years ago
3 0

Answer:

E =   k*Q₁/R₁² V/m

V =  k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                                        F = k*Q₁/R₁²

- Then the electric field at that point is

                                        E =  F/1

                                        E =  k*Q₁/R₁²  V/m

- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                                        V =  k*Q₁/R₁  Volt

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Use the circuit diagram to decide if the lightbulb will light. Justify your answer.
podryga [215]

Answer:

The lightbulb will NOT light.

Explanation:

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This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

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3 years ago
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If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?
Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

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hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

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5 0
3 years ago
A snail moves 40 cm in<br> 20 seconds. How fast<br> did it move?
nasty-shy [4]

Answer:

too fast to be a snail

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2 cm per second

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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax
alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0
3 years ago
Calculate the frequency of visible light having a wavelength of 410 nm
maria [59]
Given: Wavelength λ = 410 nm  convert to Meters m = 4.10 x 10⁻⁷ m

Speed of light c = 3 x 10⁸ m/s

Required: Frequency  f = ?

Formula: c = λf 

               f = c/λ

               f = 3 x 10⁸ m/s/4.10 x 10⁻⁷ m

               f = 7.32 x 10¹⁴/s  or 732 Thz (Terahertz)
6 0
3 years ago
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