Elektriniai laukai – problemos ir sprendimai

Elektriniai laukai – problemos ir sprendimai

1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both mokesčiai, what is the magnitude of the elektrinis laukas at point A?

Žinomas:

Mokestis 1 (q₁) = +Q

Mokestis 2 (q₂) = -Q

The distance between charge 1 and point A (r_1A) = ½ a

The distance between charge 2 and point A (r_2A) = ½ a

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

Ieškoma: The magnitude of the electric field

sprendimas:

Žingsnis 1.

The electric field produced by a charge +Q taške 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 taške A :

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

Elektrinio lauko atstojamoji taške A:

Žingsnis 2.

If point A is moved close to charge 1 then :

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

The distance between charge 2 and point A (r_2A) = ¾ a

The electric field produced by charge +Q taške 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.

Elektrinio lauko atstojamoji taške A:

2. Du krūviai q_A = 1 μC ir 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 ir q_B.

Žinomas:

Įkrova A (q_A) = 1 μC = 1 x 10^-6 C

Įkrova B (q_B) = 4 μC = 4 x 10^-6 C

k = 9 x 10⁹ N m² C^-2

Atstumas tarp krūvių A ir B (r_AB) = 4 cm = 0.04 metro

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

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

Žinomas: The magnitude of the electric field

sprendimas:

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 ir E_B have the opposite direction.

E = E_B - E_A = 9 x 10^{7 -} X 2.25 10⁷ = 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)

Sprendimas

If point P located at the left of Q₁; the electric field produced by Q₁ on point P points to leftward (nuo 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.

Žinomas:

Q₁ = +9 μC = +9 x 10^-6 C

Q₂ = -4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Atstumas tarp 1 ir 2 krūvių = 3 cm

Atstumas tarp Q₁ and point P (r_1P) = a

Atstumas tarp Q₂ ir taškas P (r_2P) = 3 + a

Ieškoma: Position of point P

sprendimas:

Point P located at leftward of Q₁.

The electric field produced by Q₁ taške 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 taške 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 :

Atstumas tarp Q₂ ir taškas P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

Atstumas tarp Q₁ ir taškas P (r_1P) = a = -9 cm or -1.8 cm.

Point P located at 1.2 cm 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).

sprendimas:

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 ir E₂.

Žinomas:

Charge q₁ = 5 µC = 5 x 10^-6 Kulonas

Charge q₂ = 5 µC = -5 x 10^-6 Kulonas

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

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

k = 9 x 10⁹ N m² C^-2

Ieškoma: The electric field at charge q₃

sprendimas:

The electric field 1 :

E₁ = k q₁ / r₁²

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

E₁ = (45 x 10³) / (225 × 10^-4)

E₁ = 0.2 x 10⁷ N / C

The electric field 2 :

E₂ = k q₂ / r₂²

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

E₂ = (45 x 10³) / (25 × 10^-4)

E₂ = 1.8 x 10⁷ N / C

Resultant of the electric field :

The resultant of the electric field at charge q₃ :

E = E₂ - E₁ = (1.8 x 10⁷) – (0.2 x 10⁷) = 1.6 x 10⁷ N / C

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)

Sprendimas

Žinomas:

Charge q_A = +2.5 C

Charge q_B = -2 C

Distance between charge q_A ir taškas P (r_A) = 5 m

Distance between charge q_B ir taškas P (r_B) = 2 m

k = 9 x 10⁹ N m² C^-2

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

sprendimas:

The electric field A :

E_A = k q_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 N / C

The electric field B :

E_B = k q_B / r_B²

E_B = (9 x 10⁹)(2) / (2)²

E_B = (18 x 10⁹) / 4

E_B = 4.5 x 10⁹ N / C

Resultant of the electric field :

Resultant of the electric field at point P :

E = E_B – EA = (4.5 – 0.9) x 10⁹ = 3.6 x 10⁹ N / C

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.

Žinomas:

Charge q₁ = -40 µC = -40 x 10^-6 C

Charge q₂ = +5 µC = +5 x 10^-6 C

Atstumas tarp q₁ ir taškas P (r₁) = 40 cm = 0.4 m = 4 x 10^-1 m

Atstumas tarp q₂ ir taškas P (r₂) = 10 cm = 0.1 = 1 x 10^-1 m

k = 9 x 10⁹ N m² C^-2

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

sprendimas:

The electric field 1 :

E₁ = k q₁ / r₁²

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

E₁ = (360 x 10³) / (16 × 10^-2)

E₁ = 22.5 x 10⁵ N / C

The electric field 2 :

E₂ = k q₂ / r₂²

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

E₂ = (45 x 10³) / 1 x 10^-2

E₂ = 45 x 10⁵ N / C

Resultant of the electric field :

The resultant of the electric field at point P :

E = E₂ – E1 = (45 – 22.5) x 10⁵ = 22.5 x 10⁵ N / C

E = 2.25 x 10⁶ N / C

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.

Žinomas:

1 krūvis (q₁) = -9 µC = -9.10^-6 Kulonas

2 krūvis (q₂) = 1 µC = 1.10^-6 Kulonas

Atstumas tarp q₁ ir q₂ (r₁₂) = 1 cm

k = 9.10⁹ Nm².C^-2

Ieškoma: Position of point P

sprendimas:

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

Kryptis E₁ į q₁ nes q₁ yra neigiamas.

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

Kryptis E₂ nuo q₂ nes q₂ yra teigiamas.

The electric field at point = 0.

Naudokite kvadratinę formulę:

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 = ….

Žinomas:

E_P = k Q / x²

Ieškoma: x

sprendimas:

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

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

Naudokite kvadratinę formulę:

1. Kas yra elektrinis laukas?
   • atsakymas: 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?
   • atsakymas: 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?
   • atsakymas: 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, Kur $k$ is Coulomb’s constant.
4. What is the direction of the electric field due to a positive charge?
   • atsakymas: 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?
   • atsakymas: 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?
   • atsakymas: 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?
   • atsakymas: 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?
   • atsakymas: 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?
   • atsakymas: 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 $W=q\times \Delta V$, Kur $\Delta V$ is the change in electric potential.

10. Can electric field lines ever cross each other?

    • atsakymas: 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.