PLEASE HELP
1 answer:
(a) the electric field generated by a single point charge is radial, and its direction is determined by the sign of the source charge. In particular, when the source charge is positive as in this example, the field lines are directed away from the charge (see attached figure).
(b) The electric field strength at distance r from the source charge is given by
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
The force experienced by the test charge in this electric field is equal to
where
is our test charge. Substituting, we find that the force is equal to
(c) The electric potential at a distance r from the source (positive) charge is given by
and the electric potential energy of a test charge q at distance r from the source charge is
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.
(b) The electric field strength at distance r from the source charge is given by
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
The force experienced by the test charge in this electric field is equal to
where
(c) The electric potential at a distance r from the source (positive) charge is given by
and the electric potential energy of a test charge q at distance r from the source charge is
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.
You might be interested in
The strength of the electrostatic force is inversely proportional to the square of the distance
between the charges. If the distance is doubled, the force is 1/4.
The new force is 0.80/4 = 0.20 N.
<h3>What is electrostatic force? </h3>Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.
The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.
In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:
Therefore, with q₁ and q₂ values of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is
.
To learn more about electrostatic force , visit:
#SPJ4
Answer:
The magnitude of the gravitational field strength near Earth's surface is represented by approximately .
Explanation:
Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:
Where:
- Mass of the planet Earth, measured in kilograms.
- Mass of the person, measured in kilograms.
- Radius of the Earth, measured in meters.
- Gravitational constant, measured in
.
But also, the magnitude of the gravitational field is given by the definition of weight, that is:
Where:
- Mass of the person, measured in kilograms.
- Gravity constant, measured in meters per square second.
After comparing this expression with the first one, the following equivalence is found:
Given that ,
and
, the magnitude of the gravitational field near Earth's surface is:
The magnitude of the gravitational field strength near Earth's surface is represented by approximately .