anonymous
  • anonymous
electric field at a distance from the source charge?
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
What is the question?
anonymous
  • anonymous
We can use the equation: \[\huge F=k \frac{ q_sq_t }{ r^2 }\]Where q_s is the source charge and q_t is some test charge that is a distance r from the source charge. We should also know that\[\huge F=q_tE \rightarrow E=\frac{F}{q_t}\]Combining the two equations gives us\[\huge E=k \frac{q_s}{r^2}\]Where q_s is the source charge and r is the distance from the source charge to some other test charge. NOTICE that no matter what the charge is on our test charge, the field ONLY DEPENDS ON THE SOURCE CHARGE! And of course the distance in between! It seems very counter intuitive, but an easy way to think of it is that imagine you're feeling the warmth sun. No matter how warm you are, the heat that the sun radiates and that you experience will still be the same at that distance.
anonymous
  • anonymous
@IrishBoy123 This is correct, yeah? I'm still taking electromagnetism and it's my first year >_<

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IrishBoy123
  • IrishBoy123
Coulomb's Law \[\huge F=k \frac{ q_sq_t }{ r^2 }\]
IrishBoy123
  • IrishBoy123
Newtons Law of gravitation: \[\Large F = G\dfrac{m_1m_2}{r^2}\] 🤑
anonymous
  • anonymous
I, too, made that observation when we first learned Coulomb's Law! Very interesting. . I would be intrigued to read about why there is this similarity between mass, charge, and the radius in between them..
anonymous
  • anonymous
radius between the masses and the radius between the charges, I mean. Not between the mass and charge.
IrishBoy123
  • IrishBoy123
spheres! the surface area of a sphere = \(4 \pi r^2\) if "stuff" expands radially throughout the universe from a point source, its "intensity" at radius R will be \(\propto \dfrac{1}{R^2}\)

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