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pogonyaev
3 years ago
12

Which condition is required for Coulomb's law to hold true?

Physics
1 answer:
AleksAgata [21]3 years ago
7 0
The correct answer is:
<span>Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
</span>F= \frac{1}{4 \pi \epsilon_0}  \frac{q_1 q_2}{r^2}
<span>where
</span>\epsilon_0 is the permittivity of free space
<span>q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
</span>F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}
where \epsilon_r is the relative permittivity, which takes into account the dielectric effects of the material.
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Answer:

Explanation:

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A particle beam is made up of many protons, each with a kinetic energy of 3.25times 10-15 J. A proton has a mass of 1.673 times
ArbitrLikvidat [17]

Answer:

The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x 10^{-4} N/C

Explanation:

given information,

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proton's mass, m = 1.673 x 10^{-27} kg

charge, q = 1.602 x 10^{-19} C

distance, d = 2 m

to find the electric field that will stop the proton, we can use the following equation:

E = F/q

   = (KE/d) / q ,        KE = Fd --> F = KE/d

   = KE/qd

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    = 1.01 x 10^{-4} N/C

3 0
3 years ago
Two marbles (m = 0.050 kg each) moving on a table TOWARDS each other with the speed of 0.3 m/s collide, and start moving away fr
snow_tiger [21]

Answer:

V1 = ____0.3______ m/s, V2 = ____-0.3______ m/s,

p1 = ____0.015______ kg m/s, p2 = ___-0.015_______ kg m/s,

ptotal = _0___________ kg m/s

Explanation:

Taking the marble ball that moves from right to left as Marble 1 and the one that moves from left to right as Marble 2. We take direction from left to right as negative and from right to left as positive. Therefore, velocity of marble 1 will be +0.3 m/s while velocity of marble 2 will be -0.3 m/s. The mass of marble 1, m2 0.05 Kg and same to marble 2, m2=0.05 Kg

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The total momentum before collision will be the sum of P1 and P2 hence 0.015 Kg m/s+-0.015 Kg m/s=0

Ptotal=0 Kg m/s

7 0
3 years ago
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