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whitty
20.02.2020 •
Engineering
A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at point a, which is on the x axis at x = 40.0 cm ? (b) What is the potential difference V b − V a when point b is at (40.0 cm, 30.0 cm)? (c) How much work is required to move an electron at rest from point a to rest at point b?
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Ответ:
a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work:![4.3\cdot 10^{-15} J](/tpl/images/0516/4452/85975.png)
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
where
q is the charge
r is the distance from the charge
In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.
Charge 1 is
and is located at the origin (x=0, y=0)
Charge 2 is
and is located at (x=0, y = 0.40 m)
Point A is located at (x = 0.40 m, y = 0)
The distance of point A from charge 1 is
So the potential due to charge 2 is
The distance of point A from charge 2 is
So the potential due to charge 1 is
Therefore, the net potential at point A is
b)
Here we have to calculate the net potential at point B, located at
(x = 0.40 m, y = 0.30 m)
The distance of charge 1 from point B is
So the potential due to charge 1 at point B is
The distance of charge 2 from point B is
So the potential due to charge 2 at point B is
Therefore, the net potential at point B is
So the potential difference is
c)
The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by
where
q is the charge of the particle
In this problem, we have:
Therefore, the work required on the electron is
Ответ: