Answer:
the force of attraction between the two charges is 4.2 x 10⁹ N.
Explanation:
Given;
the magnitude of first charge, q₁ = 0.06 C
the magnitude of the second charge, q₂ = 0.07 C
distance between the two charges, r = 3 m
The force of attraction between the two charges is calculated as ;
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/C²
Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.
Answer:
a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C
Explanation:
a) r = 0.76R
As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:
Q = ρ*V(r) = ρ*
where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³
Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:
E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -1.27*10³ N/C
b) r= 3.90 R
As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0
c) r = 2.8 R
As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.
First we need to find the total charge of the sphere, as follows:
Q = ρ*V =
In the same way that for a) we can write the following expression:
E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -0.21*10³ N/C
d) r= 7.30 R
In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.
As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.
This leaves an excess charge on the outer surface of the shell as follows:
Qsh = 66.7 nC - 16.2 nC = 50.5 nC
Now, we can repeat the same process than for a) and c) as follows:
E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = 0.1*10⁻³ N/C
Question is missing:
"What is the gravitational force between the Sun and Jupiter?"
Answer:
Explanation:
The gravitational force between two objects is given by
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the objects
In this problem, we have
is the mass of the sun
is the mass of Jupiter
is their separation
Solving the equation, we find