ਬਿਜਲੀ ਖੇਤਰ - ਸਮੱਸਿਆਵਾਂ ਅਤੇ ਹੱਲ

ਬਿਜਲੀ ਖੇਤਰ - ਸਮੱਸਿਆਵਾਂ ਅਤੇ ਹੱਲ

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 ਖਰਚੇ, what is the magnitude of the ਬਿਜਲੀ ਖੇਤਰ at point A?

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

ਚਾਰਜ 1 (q₁) = +Q

ਚਾਰਜ 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

ਲੋੜੀਂਦਾ: ਬਿਜਲੀ ਖੇਤਰ ਦੀ ਮਾਤਰਾ

ਹੱਲ:

XNUMX ਵਿੱਚੋਂ 1.

The electric field produced by a charge +Q ਬਿੰਦੂ 'ਤੇ 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 ਬਿੰਦੂ 'ਤੇ A :

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

ਬਿੰਦੂ A 'ਤੇ ਬਿਜਲੀ ਖੇਤਰ ਦਾ ਨਤੀਜਾ:

2 ਕਦਮ.

If point A is moved close to charge 1 then :

The distance between charge 1 and point ਏ (ਆਰ_1A) = ¼ a

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

The electric field produced by charge +Q ਬਿੰਦੂ 'ਤੇ 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.

ਬਿੰਦੂ A 'ਤੇ ਬਿਜਲੀ ਖੇਤਰ ਦਾ ਨਤੀਜਾ:

2. ਦੋ ਚਾਰਜ q_A = 1 μC ਅਤੇ q_B = 4 μC are separated by a distance of 4 cm (k = 9 x 10⁹ ਐਨ ਐਮ² C^-2). What is the magnitude of the electric field at the center between q_A ਅਤੇ q_B.

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

ਚਾਰਜ A (q_A) = 1 μC = 1 x 10^-6 C

ਚਾਰਜ B (q_B) = 4 μC = 4 x 10^-6 C

k = 9 x 10⁹ ਐਨ ਐਮ² C^-2

ਚਾਰਜ A ਅਤੇ B ਵਿਚਕਾਰ ਦੂਰੀ (r_AB) = 4 cm = 0.04 meters

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

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

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ: ਬਿਜਲੀ ਖੇਤਰ ਦੀ ਮਾਤਰਾ

ਹੱਲ:

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 ਅਤੇ ਈ_B have the opposite direction.

ਈ = ਈ_B - ਈ_A = 9x10^{7 -} 2.25 X 10⁷ = 6.75x10⁷ 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)

ਦਾ ਹੱਲ

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

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

Q₁ = +9 μC = +9 x 10^-6 C

Q₂ =-4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

ਚਾਰਜ 1 ਅਤੇ ਚਾਰਜ 2 ਵਿਚਕਾਰ ਦੂਰੀ = 3 ਸੈ.ਮੀ.

ਵਿਚਕਾਰ ਦੂਰੀ Q₁ and point P (r_1P) = ਏ

ਵਿਚਕਾਰ ਦੂਰੀ Q₂ ਅਤੇ ਬਿੰਦੂ P (r_2P) = 3 + a

ਲੋੜੀਂਦਾ: Position of point P

ਹੱਲ:

Point P located at leftward of Q₁.

The electric field produced by Q₁ ਬਿੰਦੂ 'ਤੇ ਪੀ:

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

The electric field produced by Q₂ ਬਿੰਦੂ 'ਤੇ 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 :

ਵਿਚਕਾਰ ਦੂਰੀ Q₂ ਅਤੇ ਬਿੰਦੂ P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

ਵਿਚਕਾਰ ਦੂਰੀ Q₁ ਅਤੇ ਬਿੰਦੂ P (r_1P) = a = -9 cm or -1.8 cm.

Point P located at 1.2 ਸੈ 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).

ਹੱਲ:

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 ਅਤੇ ਈ₂.

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

Charge q₁ = 5 µC = 5 x 10^-6 ਕੋਲਮਬ

Charge q₂ = 5 µC = -5 x 10^-6 ਕੋਲਮਬ

Distance between charge q₁ and charge q₃ (r₁) = 15 ਸੈਮੀ = 0.15 ਮੀਟਰ = 15 x 10^-2 ਮੀਟਰ

Distance between charge q₂ and charge q₃ (r₂) = 5 ਸੈਮੀ = 0.05 ਮੀਟਰ = 5 x 10^-2 ਮੀਟਰ

k = 9 x 10⁹ ਐਨ ਐਮ² C^-2

ਲੋੜੀਂਦਾ: The electric field at charge q₃

ਹੱਲ:

The electric field 1 :

E₁ = k q₁ / ਆਰ₁²

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

E₁ = (45 x 10³) / (225 x 10^-4)

E₁ = 0.2x10⁷ N / C

The electric field 2 :

E₂ = k q₂ / ਆਰ₂²

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

E₂ = (45 x 10³) / (25 x 10^-4)

E₂ = 1.8x10⁷ N / C

Resultant of the electric field :

The resultant of the electric field at charge q₃ :

ਈ = ਈ₂ - ਈ₁ = (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 ਐਨ ਐਮ² C^-2)

ਦਾ ਹੱਲ

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

Charge q_A = +2.5 C

Charge q_B = -2 C

Distance between charge q_A ਅਤੇ ਬਿੰਦੂ P (r_A) = 5 ਮੀਟਰ

Distance between charge q_B ਅਤੇ ਬਿੰਦੂ P (r_B) = 2 ਮੀਟਰ

k = 9 x 10⁹ ਐਨ ਐਮ² C^-2

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

ਹੱਲ:

The electric field A :

E_A = k q_A / ਆਰ_A²

E_A = (9 x 10⁹)(2.5) / (5)²

E_A = (22.5 x 10⁹) / 25

E_A = 0.9x10⁹ N / C

The electric field B :

E_B = k q_B / ਆਰ_B²

E_B = (9 x 10⁹)(2) / (2)²

E_B = (18 x 10⁹) / 4

E_B = 4.5x10⁹ N / C

Resultant of the electric field :

Resultant of the electric field at point P :

ਈ = ਈ_B – EA = (4.5 – 0.9) x 10⁹ = 3.6x10⁹ 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.

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

Charge q₁ = -40 µC = -40 x 10^-6 C

Charge q₂ = +5 µC = +5 x 10^-6 C

q ਵਿਚਕਾਰ ਦੂਰੀ₁ ਅਤੇ ਬਿੰਦੂ P (r₁) = 40 ਸੈਮੀ = 0.4 ਮੀਟਰ = 4 x 10^-1 m

q ਵਿਚਕਾਰ ਦੂਰੀ₂ ਅਤੇ ਬਿੰਦੂ P (r₂) = 10 cm = 0.1 = 1 x 10^-1 m

k = 9 x 10⁹ ਐਨ ਐਮ² C^-2

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

ਹੱਲ:

The electric field 1 :

E₁ = k q₁ / ਆਰ₁²

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

E₁ = (360 x 10³) / (16 x 10^-2)

E₁ = 22.5x10⁵ N / C

The electric field 2 :

E₂ = k q₂ / ਆਰ₂²

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

E₂ = (45 x 10³) / 1 x 10^-2

E₂ = 45x10⁵ N / C

Resultant of the electric field :

The resultant of the electric field at point P :

ਈ = ਈ₂ – E1 = (45 – 22.5) x 10⁵ = 22.5x10⁵ N / C

ਈ = 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.

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

ਚਾਰਜ 1 (q₁) = -9 µC = -9.10^-6 ਕੋਲਮਬ

ਚਾਰਜ 2 (q₂) = 1 µC = 1.10^-6 ਕੋਲਮਬ

q ਵਿਚਕਾਰ ਦੂਰੀ₁ ਅਤੇ q₂ (r₁₂) = 1 ਸੈ.ਮੀ

ਕੇ = 9.10⁹ Nm².C^-2

ਲੋੜੀਂਦਾ: Position of point P

ਹੱਲ:

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

ਦੀ ਦਿਸ਼ਾ E₁ ਨੂੰ q₁ ਕਿਉਕਿ q₁ ਨਕਾਰਾਤਮਕ ਹੈ।

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

ਦੀ ਦਿਸ਼ਾ E₂ ਤੋਂ ਦੂਰ q₂ ਕਿਉਕਿ q₂ ਸਕਾਰਾਤਮਕ ਹੈ.

The electric field at point = 0.

ਦੋਹਰੀ ਫਾਰਮੂਲਾ ਵਰਤੋ:

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 = ….

ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:

E_P = k Q / x²

ਲੋੜੀਂਦਾ: x

ਹੱਲ:

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

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

ਦੋਹਰੀ ਫਾਰਮੂਲਾ ਵਰਤੋ:

1. What is an electric field?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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, ਜਿੱਥੇ $k$ ਕੁਲੌਂਬ ਦਾ ਸਥਿਰਾਂਕ ਹੈ।
4. What is the direction of the electric field due to a positive charge?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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?
   • ਜਵਾਬ: 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 $W=q\times \Delta V$, ਜਿੱਥੇ $\Delta V$ is the change in electric potential.

10. Can electric field lines ever cross each other?

    • ਜਵਾਬ: 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.