The same 100-cm 3 lead block is carefully submerged in a container of mercury. One cm 3 of
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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
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Answer:
, assuming that
and
are point charges.
Explanation:
Let denote the coulomb constant. Let
denote the distance between the two point charges. In this question, neither
and
depend on the value of
.
By Coulomb's Law, the magnitude of electrostatic force between and
would be:
.
Find the first and second derivative of with respect to
. (Note that
.)
First derivative:
.
Second derivative:
.
The value of the coulomb constant is greater than
. Thus, the value of the second derivative of
with respect to
would be negative for all real
.
would be convex over all
.
By the convexity of with respect to
, there would be a unique
that globally maximizes
. The first derivative of
with respect to
should be
for that particular
. In other words:
<em>.</em>
.
.
In other words, the force between the two point charges would be maximized when the charge is evenly split:
.