المجالات الكهربائية - المشاكل والحلول

المجالات الكهربائية - المشاكل والحلول

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 اسعارنا محددة من قبل وزارة العمل, what is the magnitude of the الحقل الكهربائي at point A?

معروف :

تهمة 1 (q₁) = +Q

تهمة 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

مطلوب: مقدار المجال الكهربائي

الحل:

خطوة 1.

The electric field produced by a charge +Q عند النقطة 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 عند النقطة A :

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

محصلة المجال الكهربائي عند النقطة أ:

الخطوة 2.

If point A is moved close to charge 1 then :

The distance between charge 1 and point أ (ر_1A) = ¼ a

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

The electric field produced by charge +Q عند النقطة 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 عند النقطة أ :

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

محصلة المجال الكهربائي عند النقطة أ:

2. شحنتان q_A = 1 ميكروكولوم و q_B = 4 μC are separated by a distance of 4 cm (k = 9 x 10⁹ N م² C^-2). What is the magnitude of the electric field at the center between q_A و ق_B.

معروف :

الشحنة أ (q_A) = 1 μC = 1 x 10^-6 C

الشحنة ب (q)_B) = 4 μC = 4 x 10^-6 C

ك = 9 × 10⁹ N م² C^-2

المسافة بين الشحنتين A و B (r)_AB) = 4 سم = 0.04 متر

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

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

معروف: مقدار المجال الكهربائي

الحل:

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 وهاء_B have the opposite direction.

ه = ه_B - ه_A = 9 × 10^{7 -} 2.25 س 10⁷ = 6.75 × 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 × 10⁹ Nm²C^-21 ميكروكولوم = 10^-6 C)

الحلول

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

معروف :

Q₁ = +9،XNUMX μC = +9 × 10^-6 C

Q₂ = -4 μC = -4 × 10^-6 C

ك = 9 × 10⁹ Nm²C^-2

المسافة بين الشحنة 1 والشحنة 2 = 3 سم

المسافة بين Q₁ and point P (r_1P) = أ

المسافة بين Q₂ و نقطة P (r_2P) = 3 + a

مطلوب: Position of point P

الحل:

Point P located at leftward of Q₁.

The electric field produced by Q₁ عند النقطة ف:

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

The electric field produced by Q₂ عند النقطة 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 :

المسافة بين Q₂ و نقطة P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

المسافة بين Q₁ و نقطة P (r_1P) = a = -9 cm or -1.8 cm.

Point P located at سم1.2 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).

الحل:

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 وهاء₂.

معروف :

Charge q₁ = 5 ميكروكولوم = 5 × 10^-6 الكولون وحدة قياس للكهرباء

Charge q₂ = 5 ميكروكولوم = -5 × 10^-6 الكولون وحدة قياس للكهرباء

Distance between charge q₁ and charge q₃ (r₁) = 15 سم = 0.15 م = 15 × 10^-2 متر

Distance between charge q₂ and charge q₃ (r₂) = 5 سم = 0.05 م = 5 × 10^-2 متر

ك = 9 × 10⁹ N م² C^-2

مطلوب: The electric field at charge q₃

الحل:

The electric field 1 :

E₁ = kq₁ / ص₁²

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

E₁ = (45 × 10³) / (225 × 10^-4)

E₁ = 0.2 × 10⁷ N / C

The electric field 2 :

E₂ = kq₂ / ص₂²

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

E₂ = (45 × 10³) / (25 × 10^-4)

E₂ = 1.8 × 10⁷ N / C

Resultant of the electric field :

The resultant of the electric field at charge q₃ :

ه = ه₂ - ه₁ = (1.8 × 10⁷) – (0.2 x 10⁷) = 1.6 × 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 م² C^-2)

الحلول

معروف :

Charge q_A = +2.5 درجة مئوية

Charge q_B = -2 ج

Distance between charge q_A والنقطة P (r_A) = 5 م

Distance between charge q_B والنقطة P (r_B) = 2 م

ك = 9 × 10⁹ N م² C^-2

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

الحل:

The electric field A :

E_A = kq_A / ص_A²

E_A = (9 × 10⁹)(2.5) / (5)²

E_A = (22.5 × 10⁹) / 25

E_A = 0.9 × 10⁹ N / C

The electric field B :

E_B = kq_B / ص_B²

E_B = (9 × 10⁹)(2) / (2)²

E_B = (18 × 10⁹) / 4

E_B = 4.5 × 10⁹ N / C

Resultant of the electric field :

Resultant of the electric field at point P :

ه = ه_B – EA = (4.5 – 0.9) x 10⁹ = 3.6 × 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 نانومتر².C^-2 and 1 µC = 10^-6 C),. What is the magnitude of the electric field at point P.

معروف :

Charge q₁ = -40 ميكروكولوم = -40 × 10^-6 C

Charge q₂ = +5 ميكروكولوم = +5 × 10^-6 C

المسافة بين q₁ والنقطة P (r₁) = 40 سم = 0.4 م = 4 × 10^-1 m

المسافة بين q₂ والنقطة P (r₂) = 10 سم = 0.1 = 1 × 10^-1 m

ك = 9 × 10⁹ N م² C^-2

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

الحل:

The electric field 1 :

E₁ = kq₁ / ص₁²

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

E₁ = (360 × 10³) / (16 × 10^-2)

E₁ = 22.5 × 10⁵ N / C

The electric field 2 :

E₂ = kq₂ / ص₂²

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

E₂ = (45 × 10³) / 1 × 10^-2

E₂ = 45 × 10⁵ N / C

Resultant of the electric field :

The resultant of the electric field at point P :

ه = ه₂ – E1 = (45 – 22.5) x 10⁵ = 22.5 × 10⁵ N / C

E = 2.25 × 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.

معروف :

المسؤول 1 (q₁) = -9 µC = -9.10^-6 الكولون وحدة قياس للكهرباء

المسؤول 2 (q₂) = 1 µC = 1.10^-6 الكولون وحدة قياس للكهرباء

المسافة بين q₁ و ق₂ (r₁₂) = 1 سم

ك = 9.10⁹ Nm².C^-2

مطلوب: Position of point P

الحل:

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

اتجاه E₁ إلى q₁ لان q₁ سلبي.

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

اتجاه E₂ بعيدا عن q₂ لان q₂ هو إيجابي.

The electric field at point = 0.

استخدم الصيغة التربيعية:

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 = ….

معروف :

E_P = k Q / x²

مطلوب: x

الحل:

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

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

استخدم الصيغة التربيعية:

1. ما هو المجال الكهربائي؟
   • إجابة: 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?
   • إجابة: The strength or magnitude of an electric field $E$ at a point is defined as the force $F$ experienced by a positive test charge q₀â € < placed at that point, divided by the magnitude of the test charge itself: E=â € <F/q₀â € <.
3. How does the electric field due to a point charge vary with distance?
   • إجابة: 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، حيث $k$ هو ثابت كولوم.
4. What is the direction of the electric field due to a positive charge?
   • إجابة: 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?
   • إجابة: 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?
   • إجابة: 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?
   • إجابة: 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?
   • إجابة: 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?
   • إجابة: 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 $W=q\times \Delta V$، حيث $\Delta V$ is the change in electric potential.

10. Can electric field lines ever cross each other?

    • إجابة: 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.