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sweet-ann [11.9K]
1 year ago
5

an insulating sphere with a radius of 20 cm carries a uniform volume charge density of 1.5 x 10-6 c/m3. find the magnitude of th

e electric field at a point inside the sphere that lies 8.0 cm from the center.
Physics
1 answer:
adelina 88 [10]1 year ago
4 0

The magnitude of the electric field at a point inside the sphere that lies 8.0 cm from the center is  4.5 × 10^{3}

Given

Radius = 20 cm

Electric field at a point inside the sphere that lies 8.0 cm from the center.

ρ =  1.5 x 10-6 c/m3

E (r) = ρr / 3εο = 1.5 × 10^{-6} × 8 × 10^{-2}/ 3 × 8.85 × 10^{-12}  = 4.5 × 10^{3}

The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also referred to as an E-field[1]). [2] It also describes the physical field of a system of charged particles. [3] Electric charges and time-varying electric currents are the basis of electric fields. A manifestation of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature, is electromagnetic fields, which also include electric and magnetic fields.

Electric fields play a significant role in many branches of physics and are used in electrical engineering. For instance, in atomic physics and chemistry, the electric field acts as the attracting force that holds the atomic nucleus and electrons together in atoms. Additionally, it is the driving force for atoms' chemical bonds.

Learn more about electric field here:

brainly.com/question/15800304

#SPJ4

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(ω₁ / ω₂) = 1.9079

Explanation:

Given

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if

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and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴    K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁²   <em>(I)</em>

then

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if

v₂ = ω₂*R₂

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I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴    K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²   <em>(II)</em>

<em>∵   </em>K₁ = K₂    

⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒  (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒  (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒  (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

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