Électricité statique – problèmes et solutions

Électricité statique – problèmes et solutions

Force électrique

1. Point A in an champ électrique. The magnitude of the electric field at point A = 0.5 NC^-1. If a 0.25-C charge placed at point A, then what is the force électrique exerted on the charge.

Connu :

Electric field at point A = 0.5 NC^-1

Charge électrique at point A = 0.25 C

Voulait: Force électrique

Solution:

F = q E

F = (0.25 C)(0.5 nC)^-1)

F = 0.125 N

2. Two charges 5 C and 4 C are separated by a distance of 3 meters. If Coulomb constant is 9 × 10⁹ Nm² C^-2, what is the magnitude of electric force experienced by two charges?

Connu :

Charge 1 (q₁) = 5 C

Charge 2 (q₂) = 4 C

Distance between charge 1 and charge 2 (r) = 3 meters

Coulomb’s constant (k) = 9 × 10⁹ Nm² C^-2

Recherché : The magnitude of the electric force (F)

Solution:

3. Electric charge +q₁ = 10 μC ; +q₂ = 20 μC ; and q₃ are separated, as shown in figure below. The electric force exerted on charge q₂ = 0, then what is the charge of q₃.

Connu :

Charge 1 (q₁) = 10 μC = 10 x 10^-6 C

Charge 2 (q₂) = 20 μC = 20 x 10^-6 C

Recherché : What is the charge of q₃

Solution:

There are two forces acts on +q₂.

The first force is the repulsive force between charge +q₁ and charge +q₂ that is F₁₂, rightward.

The net electric force acts on q₂ = 0 then q₃ must negative.

The second force is the attractive force between charge +q₂ and -q₃ that is F₂₃, leftward.

Both of these forces act on q₂, have the same magnitude but the opposite direction.

Net force acts on +q₂ = 0.

4. Three charges q₁, q, and q₂ are in a line. If q = 5.0 μC and d = 30 cm, then what is the magnitude and direction of the electric force acts on the charge q.

Connu :

Charge 1 (q₁) = 30 μC = 30 x 10^-6 C

Charge 2 (q₂) = 60 μC = 60 x 10^-6 C

Charge 3 (q) = 5 μC = 5 x 10^-6 C

Distance between q₁ and q = d

Distance between q₂ and q = 2d

d = 30 cm = 0.3 meters

d² = (0.3)² = 0.09

Coulomb’s constant (k) = 9 x 10⁹ Nm² C^-2

Recherché : The magnitude and the direction of the electric force acts on the electric charge

Solution:

There are two forces act on q that is F₁ rightward (q and q₁ are positive so F₁ away from q and q₁) et F₂ leftward (q and q₂ are positive so F₂ away from q and q₂). First calculated F₁ Et F₂.

The net force is 7.5 Newton. The direction of the net force = the direction of F₁, rightward point to charge q₂.

Champ électrique

5. Un point charge q is at the point P in the electric field produit by the charge (+) so that it experiences a force of 0.05 N in the direction towards the charge. If the magnitude of the electric champ at point P is 2 x 10^-2 NC^-1, puis ce qui est the magnitude and type of charge that causes the électrique champ.

Connu :

Electric force (F) = 0.05 N

Electric field (E) = 2 x 10 ^-2 NC ^-1 = 0.02 NC ^-1

Voulait: The magnitude and type of electric charge

Solution:

The electric charge is calculated using a formula that expresses the relationship between electric force (F), the electric field (E) and electric charge (q):

F = q E

q = F / E = 0.05 N / 0.02 Nc ^-1 = 2.5 coulombs

The charge q experiences an electric force in the direction to charge (+) which produces an electric field, so the charge q is negative.

6. Charge A and B are separated by a distance of 4 meters. Point C is between both charges, 1 meter from point A. If Q_A = –300 μC, Q_B = 600 μC. 1/4 π ε₀ = 9 × 10⁹ Nm² C– 2, then what is the magnitude of the electric field at point C produced by both charges.

Connu :

Distance between charge A and B (r_AB) = 4 mètres

Distance between point C and charge A (r_AC) = 1 mètres

Distance between point C and charge B (r_BC) = 3 mètres

Charge A (q_A) = –300 μC = -300 x 10^-6 C = -3 x 10^-4 Coulomb

Charge B (q_B) = 600 μC = 600 x 10^-6 C = 6 x 10^-4 Coulomb

Constant (k) = 9 × 10⁹ Nm² C^-2

Recherché : Electric field at point C

Solution:

Electric field produced by charge A at point C :

Charger A is negative so that the direction of the electric field point to charge A and away from charge B (to left).

The electric field produced by charge B on point C :

Charge B is positive so that the direction of electric field away from charge B and point to charge A (to left).

The resultant of the electric field at point A :

E = E_A + _B

E = (27 x 10⁵) + (6 × 10⁵)

E = 33 x 10⁵ N / C

The direction of the electric field point to charge A and away from charge B (to left).

7. A 1-mg dust float in the air. If the charge of the dust is 0.5 μC and acceleration due to gravity is 10 m/s2, determine the magnitude of the electric field that supports dust.

Connu :

Mass of dust (m) = 1 mg = 1 x 10^-6 kg

Charge of dust (q) = 0.5 μC = 0.5 x 10^-6 C

Accélération due à la gravité (g) = 10 m/s²

Recherché : Champ électrique

Solution:

w = mg

w = weight of dust, m = mass of dust, g = acceleration due to gravity

The force of gravity acts on dust :

w = mg = (1 x 10^-6 kg)(10 m/s²) = 10 × 10^-6 kg m/s² = 10 x 10^-6 newton

The equation of the electric field :

E = F/q

E = electric field, F = electric force, q = electric charge

Substitute F in the equation of electric field with w in the equation of weight :

E = F/q = w/q

E = (10 x 10^-6 N) / (0.5 x 10^-6 C)

E = 10 N / 0,5 C

E = 20 N/C

8. Two charges q₁ = 32 μC and q₂ = -214 μC are separated by a distance of x, as shown in figure below. No electric field at point P located at 10 cm from q₂. Find x.

Connu :

Charge 1 (Q₁) = 32 μC

Charge 2 (Q₂) = -214 μC

Distance of point P from q₁ = x + 10 cm

The distance of point P from q₂ = 10cm

Voulait: x

Solution:

E₁ = electric field produced by charge Q₁. The direction of the electric field away from Q₁ because Q₁ is positive. E₂ = electric field produced by Q₂. The direction of the electric field point to Q₂ because Q₂ est négatif.

No net force at point p located at 10 cm from Q₂

Utilisez la formule quadratique :

9. A charged point q is at the point P in the electric field produit by the charge (+), thus having a force of 0.05 N. If the charge is +5 × l0^-6 Coulomb, then what is the magnitude of the electric field at point C.

Connu :

Electric force (F) = 0.05 Newton

Electric charge (Q) = +5 × l0^-6 Coulomb = 0.000005

Recherché : Electric field at point P

Solution:

E = F / Q

E = 0.05 Newton / 0.000005 Coulomb

E = 5 Newton / 0.0005 Coulomb

E = 10 000 Newton/Coulomb

E = 10⁴ N / C

E = 10⁴ NC^-1

1. Qu'est-ce que l'électricité statique ? Compagnie de Solution: Static electricity refers to the buildup of electric charge on the surface of objects. It’s called “static” because the charges remain in one area rather than moving or flowing as in current electricity.
2. How is static electricity generated? Compagnie de Solution: Static electricity is often generated through the process of triboelectric charging, where certain materials become charged when they come into frictional contact with a different material. Rubbing a balloon against hair or walking across a carpet in rubber-soled shoes can lead to the buildup of static charges.
3. Why do you sometimes get a shock when touching a doorknob after walking on a carpet? Compagnie de Solution: Walking on a carpet, especially in dry conditions, can cause a buildup of static charges on your body. When you touch a doorknob or another conductor, the charges can discharge quickly, resulting in a static shock.
4. How does humidity affect static electricity? Compagnie de Solution: Humidity tends to reduce the buildup of static electricity. The water molecules in the air can help dissipate electric charges. On drier days, there’s a higher likelihood of experiencing static buildup and shocks.
5. Why are some materials more prone to developing a static charge than others? Compagnie de Solution: Materials differ in their ability to hold or transfer electrons. Some materials, like rubber or certain plastics, tend to gain or lose electrons more easily than others, making them more prone to static charge buildup.
6. What is the principle behind an electroscope? Compagnie de Solution: An electroscope is a device used to detect the presence of a static charge. It typically consists of a metal rod connected to two thin leaves of metal foil. When a charged object is brought near the rod, the leaves move apart due to the repulsion of like charges, indicating the presence of static electricity.
7. How does grounding eliminate static charges? Compagnie de Solution: Grounding provides a path for excess charge to move to the Earth, which is a vast reservoir of charges. When a statically charged object is grounded, the extra charges are neutralized, reducing or eliminating the static buildup.
8. Why might you see sparks in a grain silo or during the refueling of an airplane, and why is this dangerous? Compagnie de Solution: Grain moving or rubbing against surfaces can generate static electricity, and the same goes for the flow of fuel. If there’s a buildup of static charge, it can discharge as a spark. In environments with flammable gases, dust, or liquids (like in a grain silo or during airplane refueling), these sparks can ignite the flammable material, leading to explosions.
9. How do antistatic sprays or products work? Compagnie de Solution: Antistatic sprays and products typically contain substances that make surfaces slightly conductive. By doing so, they prevent the buildup of static charges or help dissipate any charges that do build up.

10. What is the triboelectric series? Compagnie de Solution: The triboelectric series is a list that ranks materials based on their tendency to become positively or negatively charged when rubbed against another material. Materials at the top of the list tend to become positively charged, while those at the bottom tend to become negatively charged.