A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center
1 answer:
Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) We can apply Gauss's law for this problem,
Ф = ∫ E. dA = /ε₀
By taking a spherical Gaussian surface, the electric field lines are parallel to the sphere's radius, simplifying the scalar product to an algebraic one.
The surface area of a sphere is given by:
A = 4π r²
Using the density formula:
ρ = q_{int} / V
q_{int} = ρ V
The sphere's volume is calculated as:
V = 4/3 π r³
Substituting in, we get:
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
The density then is:
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
Let’s calculate for:
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / C
When r = 6.00 cm:
the Gaussian surface envelops the sphere entirely, capturing all the charge:
E (4π r²) = Q /ε₀
E = k q / r²
Calculating again:
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) Now, let’s repeat the calculations considering a conducting sphere.
For r = 4 cm:
All charge is on the sphere's surface due to the repulsion of charges, resulting in zero field inside:
E = 0
For r = 0.06 m, all charge is contained within the Gaussian surface, thus:
E = k q / r²
E = 7.49 10³ N / C
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Answer:
0.0984
Explanation:
The first diagram below illustrates a free body diagram that will aid in resolving this problem.
According to the diagram, the force's horizontal component can be expressed as:
Substituting 42° for θ and 87.0° for
Meanwhile, the vertical component is:
Again substituting 42° for θ and 87.0° for
In resolving the vector, let A denote the components in mutually perpendicular directions.
The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ
The frictional force can be expressed as:
Where;
is the coefficient of friction
N = the normal force
Also, the normal reaction (N) is calculated as mg - F sin θ
Substituting . Normal reaction becomes:
By balancing the forces, the horizontal component of the force equals the frictional force.
The horizontal component is described as follows:
Rearranging the equation above to isolate leads to:
Substituting in the following values:
m = 73 kg
g = 9.8 m/s²
Thus:
Therefore, the coefficient of friction is = 0.0984
