Answer:
A ball moving until gravity pulls it back down to the ground
Explanation:
Answer:
The charge in each ball will be 3 * 10^-12 C
Explanation:
(Assuming the correct charge of the second ball is 8 * 10^-12)
When the balls are brought in contact, all the charges are split evenly among then.
So first we need to find the total charge combined:
(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C
Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:
(9 * 10^-12) / 3 = 3 * 10^-12 C
So the charge in each ball will be 3 * 10^-12 C
Go to slader.com and type in the full name of the text book and page number. It should have the andwers
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 )