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3 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

Physics

1 answer:

il63 [147K]3 years ago

4 0

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

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A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

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Explanation:

given,

charge = -5.0 μC

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force would a proton experience = ?

we know

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$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

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Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

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2 years ago

What is the difference between delta and harbor?

Margarita [4]

The difference between delta and harbor is is that delta is the fourth letter of the modern greek alphabet while harbor is a sheltered expanse of water, adjacent to land, in which ships may dock or anchor, especially for loading and unloading.

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3 years ago

Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

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1 year ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

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