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laiz [17]
3 years ago
10

Why can you see your own reflected image in a mirror but not on a dry, painted wall?

Physics
1 answer:
kaheart [24]3 years ago
7 0
A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall. 
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Two like-charged particles are placed close to each other. How would the force of repulsion be affected if the charge on one of
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Answer: It will remain the same

Explanation:

According to <u>Coulomb's Law:</u>    

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".  </em>

<em />

Mathematically this law is written as:  

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:  

F_{E}  is the electrostatic force  

K is the Coulomb's constant  

q_{1}=q_{2}=q are the electric charges , which in this case have the same positive charge

d is the separation distance between the charges

Rewritting we have:

F_{E}=K\frac{q^{2}}{d^{2}}    (2)

Now, if the first charge is doubled:

q_{1}=2q_

And the second is reduced to a half:

q_{2}=\frac{1}{2}q

We will have the following:

F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}} (3)

F_{E}=K\frac{q^{2}}{d^{2}} (4)

As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same

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