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 )
No because you don’t learn about synthetic inventions yet in your first year
Answer:
4. F
5. F
Explanation:
jupiter constantly has hurricanes and uranus has storms often too.
lightning can move many ways including from cloud to cloud and even from ground ton sky (seriously look it up it's cool)
<span>the smallest structural and functional unit of an organism, typically microscopic and consisting of cytoplasm and a nucleus enclosed in a membrane. Microscopic organisms typically consist of a single cell, which is either eukaryotic or prokaryotic.</span>
Answer:
Mass – The single most important property that determines other properties of the star. Luminosity – The total amount of energy (light) that a star emits into space. Temperature – surface temperature, closely related to the luminosity and color of the star
Explanation:
