Electric Field Example

Electric field: It is called the space that forms an electrically charged body.
Field produced by a point charge: It is called the space where a charge that can be imagined concentrated in a point produces an electric field.
Lines of force: They are those lines drawn in an electric field of an electrically charged area in such a way that they are tangent to it at each point. The following properties of lines of force are worth noting:
There is no line of force that begins or ends in the space surrounding the charges.
All lines of force diverge radially from the positive charges, while they converge radially toward the negative charges.
Lines of force never cross.
Electric field intensity: It is equal to the quotient of dividing the force (F) received by the test load by its value (q₂), when the test load is placed on the point considered.

F = N
what₂= C
E = N / C
k = 9x10⁹ Nm²/ C²
r = m

E = F / q²= (k) q²/ r²

EXAMPLE OF APPLICATION PROBLEM:
Calculate the electric field intensity of a 3x10 test charge^-8 C receiving a force of 7.2x1 0^-10 N.

E = F / q₂= 7.2x10^-10 N / 3x10^-8 C = 2.4 x 10^-2 N / C

Electric potential energy: It is called the work necessary to move a charge from one point to another within an electric field.
Electric potential: It is the work that is carried out within an electric field to make electric charges pass within the system.
Potential at one point: It is the work supplied on the unit of charge to bring it from a point of zero potential energy to the point considered. The units of electric potential are: Volts (V).

W = J
V = V
q = C

V = W / q = q / r

EXAMPLE OF APPLICATION PROBLEM:

Calculate the work supplied at a 3x10 electrical load^-8 C which produces 2.4x10² V.
W = Vq = (2.6x10² V) (3x10-¹⁰ C) = 7.8x10^-8J
Concept of the potential difference between any two points, produced by a point charge: It is the work supplied to bring the point load unit from a zero point (a) to a second considered point (b).

V = V
W = J
q = C

V_b-V_to= W_ab/ q

V = W / q

EXAMPLE OF APPLICATION PROBLEM:

Calculate the potential difference that is generated between a point charge of 6.2x10^-9 C moving with a force of 67 N and the distance is 2 cm.
In this type of problem you have to calculate the work.

W = Fd = (67 N) (0.02 m) = 1.34 J
The potential difference is determined taking into account that the charge starts from a zero point.

V = W / q = 1.34 J / 6.2x10^-9 C = 2.16x10⁸ V

Work in an electric field: It is the amount of electric potential energy found in the electric field.

W = J
V = V
q = C

W = q (V_b-V_to)
W = qV

EXAMPLE OF APPLICATION PROBLEM:

Calculate the charge that is located in an electric field in which the work is equal to 7.9 J and the potential difference of the initial point of 2.4x1O⁸ V and the end point is 5.3x10⁸ V.

W = q (V_b-V_to)
q = W / (V_b-V_to) = 7.9 J / 5.3 x 10⁸ V-2.4x10⁸ V = 2.72x10^-8 C

Electric current: It is the product of the potential difference between two points of a conductor caused by a generator of the movement of electrons through a conductor.
Forward current (cd): It is called the electric current in which the electrons circulate in one direction (-) to (+), they also have constant intensity and the most common examples are: batteries and accumulators.
Alternating current (ac): It is called the electric current where the electrons change their sense a certain number of times, in addition the intensity is not constant and the most common examples are generators.
Electric current intensity: It is called the electric charge that passes through each section of the conductor in one second. Its units are amps (A).

l = A
q = C
t = s

l = q / t

EXAMPLE OF APPLICATION PROBLEM:
Calculate the intensity of the electric current that generates a 2.6x10 charge^-8 C in a time of 25s.

l = q / t = 2.6x10^-8 C / 25 s = 1.04x10^-9 TO

Voltage: It is the potential difference that results from the product of the resistance of the circuit and the intensity of the current. Its units are volts (V).

V = V
l = A
R = W

V = Rl

EXAMPLE OF APPLICATION PROBLEM:

Calculate the voltage generated by an electrical circuit that has a resistance of 8 Ohms and a current of 12 A.

V = Rl = (8 O) (12 A) = 96 V

Endurance: It is the opposition to the passage of an electric current within a system.
Units: Ohms (O).
R = O
V = V
l = A

R = V / l

EXAMPLE OF APPLICATION PROBLEM:
Calculate the resistance of a circuit that has a voltage of 48 V and a load of 12x10^-8 C in 3x10^-4 s.

l = q / t = 12x10^-8 C / 3x10^-4 s = 4x10^-4 TO
R = V / l = 48 V / 4x10^-4 A = 12 Ohms