Campos eléctricos: problemas y soluciones.

Campos eléctricos: problemas y soluciones.

1. El punto A se encuentra en el centro entre dos cargas. Ambas cargas tienen la misma magnitud pero signo opuesto y están separadas por una distancia a. La magnitud del campo eléctrico en el punto A es de 36 N/C. Si el punto A se acerca 1/2a a una de las cargas , ¿cuál es la magnitud del campo eléctrico en el punto A?

Conocido :

Carga 1 (q ₁ ) = +Q

Carga 2 (q ₂ ) = -Q

La distancia entre la carga 1 y el punto A (r _1A ) = ½ a

La distancia entre la carga 2 y el punto A (r _2A ) = ½ a

The magnitude of the electric field at point A (E_A) = 36 NC^-1

Querido: La magnitud del campo eléctrico

solución:

Paso 1.

The electric field produced by a charge +Q en el punto A :

Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric charge produced by a charge -Q en el punto A :

Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

La resultante del campo eléctrico en el punto A:

Paso 2:

If point A is moved close to charge 1 then :

The distance between charge 1 and point Un (r_1A) = ¼ a

La distancia entre la carga 2 y el punto A (r_2A) = ¾ a

The electric field produced by charge +Q en el punto A :

Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric field produced by charge -Q at point A :

Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

La resultante del campo eléctrico en el punto A:

2. Dos cargas q_A = 1 μC y q_B = 4 μC are separated by a distance of 4 cm (k = 9 x 10⁹ N·m² C^-2). What is the magnitude of the electric field at the center between q_A yq_B.

Conocido :

Carga A (q_A) = 1 μC = 1 x 10^-6 C

Carga B (q_B) = 4 μC = 4 x 10^-6 C

k = 9 x 10⁹ N·m² C^-2

Distancia entre las cargas A y B (r_AB) = 4 cm = 0.04 metros

Distance between charge A and the center point (r_A) = 0.02 metros

Distance between charge B and the center point (r_B) = 0.02 metros

Conocido: La magnitud del campo eléctrico

solución:

The electric field produced by charge A at the center point :

Test charge is positive and charges A is positive so that the direction of the electric field points to charge B.

The electric field produced by charge B at the center point :

Test charge is positive and charge B is positive so that the direction of the electric field points to charge A.

The resultant of the electric field at the center point :

E_A y E_B have the opposite direction.

mi = mi_B - E_A = 9 x 10^{7 -} 2.25 x 10⁷ = 6.75 x 10⁷ NC^-1

3. According to figure below, where the point P is located so that the magnitude of the electric field at point P = 0 ? (k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

Solución

If point P located at the left of Q₁; the electric field produced by Q₁ on point P points to leftward (lejos de Q₁) and the electric field produced by Q₂ on point P points to rightward (point to Q₁). The direction of the electric field is opposite so that the electric field at point P = 0.

Conocido :

Q ₁ = +9 μC = +9 x 10 ⁻⁶ C

Q ₂ = -4 μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Distancia entre la carga 1 y la carga 2 = 3 cm

Distancia entre Q₁ and point P (r_1P) = un

Distancia entre Q₂ y punto P (r_2P) = 3 + a

Buscado : Position of point P

solución:

Point P located at leftward of Q₁.

The electric field produced by Q₁ en el punto P:

Test charge is positive and Q₁ is positive so that the direction of the electric field to leftward.

The electric field produced by Q₂ en el punto P :

Test charge is positive and Q₂ is negative so that the direction of the electric field to rightward.

Resultant of the electric field at point A :

Use quadratic formula to find a :

Distancia entre Q₂ y punto P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

Distancia entre Q₁ y punto P (r_1P) = a = -9 cm or -1.8 cm.

Point P located at 1.2cm rightward of Q₂.

4. Charge q₃ located at 5 cm rightward of q₂, as shown in the figure below. What is the magnitude of the electric field at charge q₃ (1 µC = 10^-6 C).

solución:

Charge q₃ is positive so that the direction of the electric field at charge q₃ points to the minus charge q₂ (E₂) and away from the plus charge q₁ (E₁). The resultant of the electric field is the sum of the electric field E₁ y E₂.

Conocido :

Charge q₁ = 5 µC = 5 x 10^-6 Culombio

Charge q₂ = 5 µC = -5 x 10^-6 Culombio

Distance between charge q₁ and charge q₃ (r₁) = 15 cm = 0.15 m = 15 x 10^-2 metros

Distance between charge q₂ and charge q₃ (r₂) = 5 cm = 0.05 m = 5 x 10^-2 metros

k = 9 x 10⁹ N·m² C^-2

Buscado : The electric field at charge q₃

solución:

The electric field 1 :

E₁ = kq₁ / r₁²

E₁ = (9 x 10⁹(5 x 10^-6) / (15 x 10^-2)²

E₁ = (45 x 10³) / (225 x 10^-4)

E₁ = 0.2 x 10⁷ CAROLINA DEL NORTE

The electric field 2 :

E₂ = kq₂ / r₂²

E₂ = (9 x 10⁹(5 x 10^-6) / (5 x 10^-2)²

E₂ = (45 x 10³) / (25 x 10^-4)

E₂ = 1.8 x 10⁷ CAROLINA DEL NORTE

Resultant of the electric field :

The resultant of the electric field at charge q₃ :

mi = mi₂ - E₁ = (1.8 x 10⁷) – (0.2 x 10⁷) = 1.6 x 10⁷ CAROLINA DEL NORTE

The direction of the electric field points to leftward (same direction as E₂).

5. Two charges are separated as shown in figure below. What is the electric field at point P (k = 9 x 10⁹ N·m² C^-2)

Solución

Conocido :

Charge q_A = +2.5 °C

Charge q_B = -2 C

Distance between charge q_A y el punto P (r_A) = 5 m

Distance between charge q_B y el punto P (r_B) = 2 m

k = 9 x 10⁹ N·m² C^-2

Buscado : the magnitude of the electric field at point P.

solución:

The electric field A :

E_A = kq_A / r_A²

E_A = (9 x 10⁹(2.5) / (5)²

E_A = (22.5 x 10⁹) / 25

E_A = 0.9 x 10⁹ CAROLINA DEL NORTE

The electric field B :

E_B = kq_B / r_B²

E_B = (9 x 10⁹(2) / (2)²

E_B = (18 x 10⁹) / 4

E_B = 4.5 x 10⁹ CAROLINA DEL NORTE

Resultant of the electric field :

Resultant of the electric field at point P :

mi = mi_B – EA = (4.5 – 0.9) x 10⁹ = 3.6 x 10⁹ CAROLINA DEL NORTE

The direction to leftward (same direction as E_B).

6. Two charges Q₁ = -40 µC and Q₂ = +5 µC as shown in figure below (k = 9 x 10⁹ nm².C^-2 and 1 µC = 10^-6 C),. What is the magnitude of the electric field at point P.

Conocido :

Charge q₁ = -40 µC = -40 x 10^-6 C

Charge q₂ = +5 µC = +5 x 10^-6 C

Distancia entre q₁ y el punto P (r₁) = 40 cm = 0.4 m = 4 x 10^-1 m

Distancia entre q₂ y el punto P (r₂) = 10 cm = 0.1 = 1 x 10^-1 m

k = 9 x 10⁹ N·m² C^-2

Buscado : the magnitude of the electric field at point P.

solución:

The electric field 1 :

E₁ = kq₁ / r₁²

E₁ = (9 x 10⁹(40 x 10^-6) / (4 x 10^-1)²

E₁ = (360 x 10³) / (16 x 10^-2)

E₁ = 22.5 x 10⁵ CAROLINA DEL NORTE

The electric field 2 :

E₂ = kq₂ / r₂²

E₂ = (9 x 10⁹(5 x 10^-6) / (1 x 10^-1)²

E₂ = (45 x 10³) / 1 x 10^-2

E₂ = 45 x 10⁵ CAROLINA DEL NORTE

Resultant of the electric field :

The resultant of the electric field at point P :

mi = mi₂ – E1 = (45 – 22.5) x 10⁵ = 22.5 x 10⁵ CAROLINA DEL NORTE

E = 2.25 x 10⁶ CAROLINA DEL NORTE

The direction of the electric field points to rightward (same direction as E₂).

7. Two point charges as shown in figure below.

Where is point P located so that the magnitude of the electric field at point P = 0. k = 9.10⁹ Nm².C^-2, 1 µC = 10^-6 C.

Conocido :

Cargo 1 (q₁) = -9 µC = -9.10^-6 Culombio

Cargo 2 (q₂) = 1 µC = 1.10^-6 Culombio

Distancia entre q₁ yq₂ (r₁₂) = 1cm

k = 9.10⁹ Nm².C^-2

Buscado : Position of point P

solución:

E₁ = the magnitude of the electric field produced by q₁ at point P

La dirección de E₁ a q₁ because q₁ es negativo

E₂ = the magnitude of the electric field produced by q₂ at point P

La dirección de E₂ lejos de q₂ because q₂ es positivo.

The electric field at point = 0.

Utilice la fórmula cuadrática:

Distance between P and q₂ = x = 0.5 cm.

Point P located at 0.5 cm rightward q₂ or 0.25 cm leftward q₁.

8. According to the figure below, if the magnitude of the electric field at point P = k Q/x², then x = ….

Conocido :

E_P = k Q / x²

Se busca: x

solución:

E₂ = The magnitude of the electric field at point P by charge +32Q

r₂ =Distance between charge +32Q and point P = a + x

Utilice la fórmula cuadrática:

1. ¿Qué es un campo eléctrico?
   • Respuesta: An electric field is a region around a charged object where electric forces can be experienced by other charged objects. It is a vector field, meaning it has both magnitude and direction at every point.
2. How is the strength of an electric field determined?
   • Respuesta: The strength or magnitude of an electric field $E$ at a point is defined as the force $F$ experienced by a positive test charge q₀​ placed at that point, divided by the magnitude of the test charge itself: E=​F/q₀​.
3. How does the electric field due to a point charge vary with distance?
   • Respuesta: The electric field $E$ due to a point charge $Q$ is inversely proportional to the square of the distance $r$ from the charge. The relationship is given by E=kQ​/r², donde $k$ es la constante de Coulomb.
4. What is the direction of the electric field due to a positive charge?
   • Respuesta: The electric field due to a positive charge points radially outward from the charge. For a negative charge, the field points radially inward, towards the charge.
5. How can you represent electric fields graphically?
   • Respuesta: Electric fields can be represented graphically using field lines (or lines of force). The direction of the field at any point is tangent to the field line at that point, and the density of the lines indicates the magnitude of the field.
6. What happens to the electric field inside a conductor in electrostatic equilibrium?
   • Respuesta: Inside a conductor in electrostatic equilibrium, the electric field is zero. This is because any external field causes free electrons in the conductor to redistribute, cancelling the external field inside.
7. How do electric field lines behave near a sharp edge of a conductor?
   • Respuesta: Near a sharp edge or pointed tip of a conductor, the electric field lines are more concentrated, leading to a stronger electric field in that region. This is the basis for the operation of devices like the lightning rod.
8. How do superposition principles apply to electric fields?
   • Respuesta: The electric field due to multiple charges at a point is simply the vector sum of the electric fields due to each individual charge. This is known as the superposition principle.
9. How is the work done by an external agent related to the electric field when moving a charge within the field?
   • Respuesta: The work $W$ done by an external agent in moving a charge $q$ from one point to another in an electric field is equal to the negative of the change in electric potential energy, which is $W=q\times \Delta V$, donde $\Delta V$ is the change in electric potential.

10. Can electric field lines ever cross each other?

    • Respuesta: No, electric field lines cannot cross each other. If they did, it would imply that at the point of intersection, there are two different directions of the electric field, which is not possible.