a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work: [tex]4.3\cdot 10^{-15} J[/tex]
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
[tex]V(r)=\frac{kq}{r}[/tex]
where
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
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
[tex]q_1=+2.00\mu C=+2.00\cdot 10^{-6}C[/tex]
and is located at the origin (x=0, y=0)
Charge 2 is
[tex]q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C[/tex]
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
[tex]r_{1A}=0.40 m[/tex]
So the potential due to charge 2 is
[tex]V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V[/tex]
The distance of point A from charge 2 is
[tex]r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m[/tex]
So the potential due to charge 1 is
[tex]V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V[/tex]
Therefore, the net potential at point A is
[tex]V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V[/tex]
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
[tex]r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m[/tex]
So the potential due to charge 1 at point B is
[tex]V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V[/tex]
The distance of charge 2 from point B is
[tex]r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m[/tex]
So the potential due to charge 2 at point B is
[tex]V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V[/tex]
Therefore, the net potential at point B is
[tex]V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V[/tex]
So the potential difference is
[tex]V_B-V_A=-29,500 V-(-2700 V)=-26,800 V[/tex]
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
[tex]W=q\Delta V[/tex]
where
q is the charge of the particle
[tex]\Delta V[/tex] is the potential difference
In this problem, we have:
[tex]q=-1.6\cdot 10^{-19}C[/tex] is the charge of the electron
[tex]\Delta V=-26,800 V[/tex] is the potential difference
Therefore, the work required on the electron is
[tex]W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J[/tex]
