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Mrac [35]
3 years ago
10

A hollow metal sphere is electrically neutral (no excess charge). A small amount of negative charge is suddenly placed at one po

int P on this metal sphere. If we check on this excess negative charge a few seconds later we will find one of the following possibilities:
a. All of the excess charge remains right around P.
b. The excess charge has distributed itself evenly over the outside surface of the sphere.
c. The excess charge is evenly distributed over the inside and outside surface.
d. Most of the charge is still at point P, but some will have spread over the sphere.
e. There will be no excess charge left.
What is the The original magnitude of the force the acharge was magnitude of the force on now?
Physics
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:

b. The excess charge has distributed itself evenly over the outside surface of the sphere.

Explanation:

When a neutral hollow metal sphere is touched externally with a charge, the charge distributes itself evenly on the surface of the sphere. One thing to note here is that for a hollow sphere, the charges stays on the surface alone. This phenomenon is why the inside of airplanes are free from electric charge when struck by a lightning when they travel through a lightning storm.

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3 years ago
A boy runs 400m at an average speed of 4.0m/s he runs the first 200m in 40 s how long does he take to run the second 200m?
IRISSAK [1]
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Est ce qu’un cricket c’est un consommateur
Sholpan [36]

Answer:

Oui

Explanation:

7 0
3 years ago
Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional
Alexxx [7]

Answer:

165.529454

Explanation:

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7 0
3 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

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

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

8 0
3 years ago
Read 2 more answers
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