Goobaha korontada - dhibaatooyinka iyo xalalka

Goobaha korontada - dhibaatooyinka iyo xalalka

1. Barta A waxay ku taal bartamaha u dhaxaysa laba dallacaad. Labada dallacaadba waxay leeyihiin cabbir isku mid ah laakiin calaamad ka soo horjeeda waxaana kala fogeeya masaafada a. Cabbirka goobta korantada ee barta A waa 36 N/C. Haddii barta A ay u dhaqaaqdo 1/2a meel u dhow mid ka mid ah labada dallacaad , waa maxay baaxadda goobta korontada ee barta A?

La yaqaan:

Kharash 1 (q ₁ ) = +Q

Kharash 2 (q ₂ ) = -Q

Masaafada u dhaxaysa kharashka 1 iyo dhibicda A (r _1A ) = ½ a

Masaafada u dhaxaysa kharashka 2 iyo dhibicda A (r _2A ) = ½ a

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

SE buska: Baaxadda goobta korantada

Xalka:

Step 1.

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

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

Natiijada garoonka korantada ee barta A:

Tallaabada 2.

If point A is moved close to charge 1 then :

The distance between charge 1 and point A (r)_1A) = ¼ a

Masaafada u dhaxaysa kharashka 2 iyo dhibicda A (r_2A) = ¾ a

The electric field produced by charge +Q barta 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.

Natiijada garoonka korantada ee barta A:

2. Laba dacwadood q_A = 1 μC iyo 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 iyo q_B.

La yaqaan:

Dalac A (q)_A) = 1 μC = 1 x 10^-6 C

Dalac B (q)_B) = 4 μC = 4 x 10^-6 C

k = 9 x 10⁹ N m² C^-2

Masaafada u dhaxaysa khidmadaha A iyo B (r)_AB) = 4 cm = 0.04 mitir

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

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

La yaqaan Baaxadda goobta korantada

Xalka:

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 iyo E_B have the opposite direction.

E = E_B - E_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)

Solution

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

La yaqaan:

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

Masaafada u dhaxaysa dallacaadda 1 iyo dallacaadda 2 = 3 cm

Fogaanta u dhaxaysa Q₁ and point P (r_1P) = a

Fogaanta u dhaxaysa Q₂ iyo dhibic P (r_2P) = 3 + a

La doonayo: Position of point P

Xalka:

Point P located at leftward of Q₁.

The electric field produced by Q₁ barta 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 barta 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 :

Fogaanta u dhaxaysa Q₂ iyo dhibic P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

Fogaanta u dhaxaysa Q₁ iyo dhibic 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).

Xalka:

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 iyo E₂.

La yaqaan:

Charge q₁ = 5 µC = 5 x 10^-6 Coulomb

Charge q₂ = 5 µC = -5 x 10^-6 Coulomb

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

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

k = 9 x 10⁹ N m² C^-2

La doonayo: The electric field at charge q₃

Xalka:

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 = 0.2x10⁷ N / C

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 = 1.8x10⁷ 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)

Solution

La yaqaan:

Charge q_A = +2.5 C

Charge q_B = -2 C

Distance between charge q_A iyo dhibic P (r)_A) = 5 m

Distance between charge q_B iyo dhibic P (r)_B) = 2 m

k = 9 x 10⁹ N m² C^-2

La doonayo: the magnitude of the electric field at point P.

Xalka:

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 = 0.9x10⁹ N / C

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 = 4.5x10⁹ 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 = 3.6x10⁹ 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.

La yaqaan:

Charge q₁ = -40 µC = -40 x 10^-6 C

Charge q₂ = +5 µC = +5 x 10^-6 C

Masaafada u dhaxaysa q₁ iyo dhibic P (r)₁) = 40 cm = 0.4 m = 4 x 10^-1 m

Masaafada u dhaxaysa q₂ iyo dhibic P (r)₂) = 10 cm = 0.1 = 1 x 10^-1 m

k = 9 x 10⁹ N m² C^-2

La doonayo: the magnitude of the electric field at point P.

Xalka:

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 = 22.5x10⁵ N / C

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 = 45x10⁵ N / C

Resultant of the electric field :

The resultant of the electric field at point P :

E = E₂ – E1 = (45 – 22.5) x 10⁵ = 22.5x10⁵ 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.

La yaqaan:

Kharash 1 (q₁) = -9 µC = -9.10^-6 Coulomb

Kharash 2 (q₂) = 1 µC = 1.10^-6 Coulomb

Masaafada u dhaxaysa q₁ iyo q₂ (r₁₂) = 1cm

k = 9.10⁹ Nm².C^-2

La doonayo: Position of point P

Xalka:

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

Jihada E₁ ilaa q₁ maxaa yeelay, q₁ waa taban.

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

Jihada E₂ ka fog q₂ maxaa yeelay, q₂ waa togan

The electric field at point = 0.

Isticmaal qaacidada labajibbaaran:

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 = ….

La yaqaan:

E_P = k Q / x²

La rabay : x

Xalka:

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

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

Isticmaal qaacidada labajibbaaran:

1. What is an electric field?
   • Jawaab: 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?
   • Jawaab: 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?
   • Jawaab: 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, halkaas oo $k$ waa joogtada Coulomb.
4. What is the direction of the electric field due to a positive charge?
   • Jawaab: 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?
   • Jawaab: 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?
   • Jawaab: 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?
   • Jawaab: 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?
   • Jawaab: 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?
   • Jawaab: 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 $W=q\times \Delta V$, halkaas oo $\Delta V$ is the change in electric potential.

10. Can electric field lines ever cross each other?

    • Jawaab: 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.