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VARVARA [1.3K]
3 years ago
15

Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a)

what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

14400\ \text{N}, Attractive

3240\ \text{N}, Repulsive

Explanation:

q_1 = -20 μC

q_2 = 50 μC

r = Distance between the charges = 2.5 cm

k = Coulomb constant = 9\times 10^9\ \text{Nm}^2/\text{C}^2

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}

The magnitude of force each sphere will experience is 14400\ \text{N}

Since the charges have opposite charges they will attract each other.

Now the charges are brought into contact with each other so the resultant charge will be

q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}

F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}

The magntude of the force the spheres experience will be 3240\ \text{N}

The spheres have the same charge now so they will repel each other.

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My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
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Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

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Answer:

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