Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
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Explanation:because she is telling you to do a project
From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
The given weight of Elliot is 600 N
From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.
P.E = mgh
where mg = Weight = 600
To find the height Elliot will reach, substitute all necessary parameters into the equation above.
250 = 600h
Make h the subject of the formula
h = 250/600
h = 0.4167 meters
Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
Learn more about energy here: brainly.com/question/24116470
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