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melamori03 [73]
4 years ago
6

A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance r

l (r (A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Physics
1 answer:
netineya [11]4 years ago
5 0

Answer:

 E ’= E / 8

therefore the correct answer is A

Explanation:

Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface

       Фi = ∫ E. dA = q_{int} /ε₀

       E (4πr²) = q_{int} / ε₀

density is

      ρ = q_{int} / V

      q_{int} = ρ V = ρ 4/3 π r³

we substitute        

        E (4π r²) = ρ 4/3 π r³ /ε₀

        E = 1 /3ε₀   ρ r

let's change the density by

       ρ = Q / V = ​​Q / (4/3 π R³)

         E = 1 / 4πε₀ Q r / R³

if we now distribute the same charge on a sphere of radius R' = 2R

        E ’= 1 / 4pieo Q r / (2R)³

        E ’= 1 / 4ft Qr / R³ ⅛

        E ’= E / 8

therefore the correct answer is A

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A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at rrr = 4.0 mm is πr
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Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

    The wavelength of the spherical wave is  \lambda =  2.0 \ mm =  \frac{2}{1000} = 0.002 \ m

    The first radius  is  r_1  = 4.0 \ mm  = \frac{4}{1000}  = 0.004 \ m

     The phase at that instant is  \phi _1 = \pi \ rad

     The second radius is  r_2  = 3.5 \ mm  = \frac{3.5}{1000}  = 0.0035 \ m

Generally the phase difference is mathematically represented as

          \Delta  \phi =  \phi _2 -  \phi _1

this can also be expressed as

         \Delta \phi =  \frac{2 \pi }{\lambda } (r_2 - r_1 )

So we have that

   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

substituting values

     \phi _2 -  \pi =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )

    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

   \phi _2  = 1.57 \  rad

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3 years ago
Can the kinetic energy of an object be negative? Can the potential energy of an object be negative? Can a potential energy funct
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Answer:

Kinetic energy cannot be negative

potential energy is a reference dependent quantity and it can be positive as well as negative both

Since potential energy is defined only for conservative force so it can not be found for friction force

Explanation:

Kinetic energy of an object is given by the formula

KE = \frac{1}{2}mv^2

here we know that

m = mass of object that can not be negative

v = speed of the object and since its square is given here so it can not be negative

so Kinetic energy is always positive

potential energy is given as the energy due to the virtue of the position of object

so it is

\Delta U = -\int F.dr

so potential energy is a reference dependent quantity and it can be positive as well as negative both

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Answer:

It can have many set of perpendicular component but the most important one is 3 AND 4

Explanation:

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Answer:

= 4.3 × 10 ⁻¹⁴ m

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Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

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=  8.459 × 10⁻¹³

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Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

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