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Gre4nikov [31]
2 years ago
11

Two small plastic spheres each have a mass of 1.1 g and a charge of -50.0 nC . They are placed 2.1 cm apart (center to center).

A. What is the magnitude of the electric force on each sphere? B. By what factor is the electric force on a sphere larger than its weight?
Physics
1 answer:
Solnce55 [7]2 years ago
3 0

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

F = \frac{kq_1q_2}{r^2}

here we will have

q_1 = q_2 = 50 nC

here the distance between the center of two balls is given as

r = 2.1 cm = 0.021 m

now we will have

F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

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Answer:

The tension is  T  = 4326.7 \  N

Explanation:

From the question we are told that

   The  total mass is  m  =  200 \  kg

    The  radius is r = 5 \  m

     The  density of air is  \rho_a  =  1.225 \ kg/m^3

Generally the upward  force acting on the balloon is mathematically represented as

        F_N  =   T  + mg

=>     (\rho_a  *  V  *  g ) =   T  + mg

=>   T  =  (\rho_a  * V  *  g   )  - mg

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=>   V  = 523.67\  m^3

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   T  = 4326.7 \  N

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