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Lostsunrise [7]
3 years ago
11

A point charge is used to determine the electric field around a charged particle. Why is it necessary that the point charge does

not affect the position of the charge being tested?
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0
The strength, and possibly the shape and direction, of the electric field
around a charged particle depends on the location of the particle. 

If the process of measuring the field causes the particle to move, then
the measurement you get wouldn't mean anything. 

Your measurements wouldn't show the ACTUAL field around the particle.
They would show what the field is like AFTER something comes along
and distorts it, and that's not what you're trying to measure.

It would be like carrying a flame thrower into a freezer when you go in
to measure the temperature in there.

Or if you had to measure how much light is leaking into a dark room,
and you carried a flashlight with you to see your way around in there.
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A solution is oversaturated with solute. which could be done to decrease the oversaturation?
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3 years ago
A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe
Marina86 [1]

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9  = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

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