Question

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?

Answer 1

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work:
`4.3\cdot 10^(-15) J`

Step-by-step explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

`V(r)=(kq)/(r)`

where

`k=8.99\cdot 10^9 Nm^(-2)C^(-2)` 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

`q_1=+2.00\mu C=+2.00\cdot 10^(-6)C`

and is located at the origin (x=0, y=0)

Charge 2 is

`q_2=-3.00 \mu C=-3.00\cdot 10^(-6)C`

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

`r_(1A)=0.40 m`

So the potential due to charge 2 is

`V_1=((8.99\cdot 10^9)(+2.00\cdot 10^(-6)))/(0.40)=+4.50\cdot 10^4 V`

The distance of point A from charge 2 is

`r_(2A)=√(0.40^2+0.40^2)=0.566 m`

So the potential due to charge 1 is

`V_2=((8.99\cdot 10^9)(-3.00\cdot 10^(-6)))/(0.566)=-4.77\cdot 10^4 V`

Therefore, the net potential at point A is

`V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V`

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

`r_(1B)=√((0.40)^2+(0.30)^2)=0.50 m`

So the potential due to charge 1 at point B is

`V_1=((8.99\cdot 10^9)(+2.00\cdot 10^(-6)))/(0.50)=+3.60\cdot 10^4 V`

The distance of charge 2 from point B is

`r_(2B)=√((0.40)^2+(0.40-0.30)^2)=0.412 m`

So the potential due to charge 2 at point B is

`V_2=((8.99\cdot 10^9)(-3.00\cdot 10^(-6)))/(0.412)=-6.55\cdot 10^4 V`

Therefore, the net potential at point B is

`V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V`

So the potential difference is

`V_B-V_A=-29,500 V-(-2700 V)=-26,800 V`

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

`W=q\Delta V`

where

q is the charge of the particle

`\Delta V` is the potential difference

In this problem, we have:

`q=-1.6\cdot 10^(-19)C` is the charge of the electron

`\Delta V=-26,800 V` is the potential difference

Therefore, the work required on the electron is

`W=(-1.6\cdot 10^(-19))(-26,800)=4.3\cdot 10^(-15) J`