Find the charge q(t) on the capacitor and the current i(t) in the given lrc-series circuit. l = 5 7 h, r = 10 ω, c = 1 70 f, e(t
) = 1400 v, q(0) = 0 c, i(0) = 0 a
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By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.
Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .
now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.
charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.
Now , use vector for solve it.
vector = vector Fe + vector Fn,
⇒ || =
⇒ Fe = Fs = KqQ/(1m)² = KqQ
⇒ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}
⇒ = √2KqQ
Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.
To learn more about positive charges here
#SPJ4
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =
λ ~ 1.97 ×10⁻⁷m
λ =
λ ~ 1.97 ×10⁻⁷m