*example included* Two uncharged spheres are separated by 3.50 m. If 1.30 ✕ 10¹² electrons are removed from one sphere and place
d on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges. Find the net charge on each sphere and substitute values into Coulomb's law.
1 answer:
Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.
<h3>Coulomb's Law</h3>Charged bodies experience a force of attraction or repulsion on approach.
From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.
From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
where:
- F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
- Q and q are the values of the two point charges. They are measured in Coulombs (C).
- d is the value of the distance that separates them. It is measured in meters (m).
- K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹
.
The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
<h3>This case</h3>In this case, you know that:
- The two uncharged sphere are separated by the distance of d= 3.50 m
- The number of electrons are 1.30×10¹².
- Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C
Replacing in Coulomb's Law:
Solving:
<u><em>F= 3.1865 N</em></u>
Finally, the magnitude of the Coulomb force is 3.1865 N.
Learn more about Coulomb's Law:
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Answer:
The potential difference between the plates increases
Explanation:
As we know that the capacitance of the capacitor is given by:
(1)
where
q = charge
C = capacitance
V = Voltage or Potential Difference
Also, the capacitance of a parallel plate capacitor is given as:
(2)
where
A = Area of the plates
D = Separation distance between the plates
Now, from eqn (1) and (2):
Now, from the above eqn we can say that:
Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.
Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.
Answer:
Explanation:
A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have
where vr is the voltage in the resistance, vi is the voltage in the inductance and vb is the source voltage. But also we have that
where L is the inductance of the circuit, r is the resistance an i is the current. By replacing we have the differential equation
I hope this is useful for you
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