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love history [14]
3 years ago
5

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

es. 4. What evidence do you see that Newton's third law applies to electrostatic forces
Physics
1 answer:
kumpel [21]3 years ago
7 0

Answer:

he combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.

Explanation:

A very simple experiment that we can do is to rub a brush or plastic comb with a piece of paper, this creates some electrical charges in the comb, we hang the comb by a thread.

We take another comb and we form it with the same paper, when we bring the combs closer, we see that they repel each other, this is proof that the electrical charges created are of the same type.  

If now we rub the comb with another material, for example plastic, and bring it closer to the comb that is hanging, we see that they are approaching, therefore, it is proven that the load created from another type

With these two simple experiments it can be proved that electric charges are of two types, which are called positive and negative.  

4. In the previous experiment we see that when the combs are brought closer, each one has to move, therefore a force must be applied to each one, and the only force is electrostatic. If we observe carefully we see that the breaking of the combs is the same, therefore the force must have the same magnitude in each one and in the opposite direction; this description is equivalent to Newton's third law.

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1,8066 x 1024 atoms

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4 years ago
A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th
sleet_krkn [62]

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

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I = Q/t

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