Answer:
Explanation:
From the question we are told that:
Coefficient of kinetic friction
Vertical Mass
Horizontal mass
Generally the equation for kinetic force is mathematically given by
Generally the equation for T is mathematically given by
For
Therefore substituting
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