How do you find the electric potential from force?

[Mitocarta]

Homework Statement

Suppose Coulomb's force is actually

F= (y)(q1q2)(r hat)/(r^4)

with y being a constant. Find the electric potential function V(x,y) for a charge Q located at the point x=a, y=b

Homework Equations

Fq=E

V=integral (E)

The Attempt at a Solution

I am very confused by this problem and do not know if my solution is correct:

1)

find electric field using fq=E,

E=(y)(q1)(r hat) / (r^4)

2)

Integrate to find V. Somehow change q1 to Q

V= integral (E dot Dr)
V= Qy integral (1/r^4)

v= -Qy/r^3,

where r = sqrt((x-A)^3 + (y-B)^2)

giving

V (x,y) = - yQ / 3 ((x-a)^2 + (y-b)^2))^(3/2)) + 0Is this correct? How does q1 change into Q?

 

[SammyS]

Homework Statement

Suppose Coulomb's force is actually

F= (y)(q1q2)(r hat)/(r^4)

with y being a constant. Find the electric potential function V(x,y) for a charge Q located at the point x=a, y=b

Homework Equations

Fq=E

V=integral (E)

The Attempt at a Solution

I am very confused by this problem and do not know if my solution is correct:

1) find electric field using fq=E,

E=(y)(q1)(r hat) / (r^4)

2) Integrate to find V. Somehow change q1 to Q

V= integral (E dot Dr)
V= Qy integral (1/r^4)

v= -Qy/r^3,

where r = sqrt((x-A)^3 + (y-B)^2)

giving

V (x,y) = - yQ / 3 ((x-a)^2 + (y-b)^2))^(3/2)) + 0

Is this correct? How does q1 change into Q?

Not correct.

Are you sure that's y, and not γ (gamma) ? Whatever it is, it's a constant.

I'm assuming that $\displaystyle \ \ \vec{F}=\gamma\,\frac{q_{1}q_{1}}{r^4} \hat{r}\ \$ gives the force exerted on q₂ by q₁ .

Then what you for the E field due to q₁ makes sense. To find the E field due to Q, simply replace q₁ with Q .To get the potential, simply integrate.

$\displaystyle V(r)=-\int \vec{E}(r)\cdot d\vec{r}$

$\displaystyle =-\int\gamma\,Qr^{-4}\hat{r}\cdot d\vec{r}$​

 

[Mitocarta]

Thank you.



I know that dr = dr ir + dθ ir and that dr = dx ix + dy iy.


How do I know in which situation to use each?

 

[SammyS]

Thank you.

I know that dr = dr ir + dθ ir and that dr = dx ix + dy iy.

How do I know in which situation to use each?

Well, $\hat{r}\ \text{ is parallel to }\ d\vec{r}\ \$ so that $\hat{r}\cdot d\vec{r}=dr\ .$

That results in $\displaystyle \int\vec{E}\cdot d\vec{r}=\int E\,dr\ .$

 

[Nickie]

To find the electric potential from force, we can use the definition of electric potential as the work done per unit charge to bring a test charge from infinity to a point in an electric field. Mathematically, it can be written as V = W/q, where V is the electric potential, W is the work done, and q is the test charge.

In this case, the force is given by F = (y)(q1q2)(r hat)/(r^4), where y is a constant, q1 and q2 are the charges, and r is the distance between them. To find the electric potential caused by a charge Q located at the point (a,b), we need to find the work done by this force to bring a test charge from infinity to the point (a,b).

The work done can be calculated by integrating the force over the distance r, which gives us W = integral (F dot dr) = integral [(y)(q1q2)(r hat)/(r^4) dot dr]. We can simplify this by noting that the direction of the force and the displacement dr are in the same direction, so we can write the dot product as F dr = F dr cosθ, where θ is the angle between the force and displacement vectors.

Since the force is radial, we can write cosθ as cosθ = r/r = 1, and the integral becomes W = integral [(y)(q1q2)(r hat)/(r^4) dr] = integral [(y)(q1q2)/(r^3) dr]. This integral can be evaluated to give W = - (y)(q1q2)/(2r^2) + C, where C is a constant of integration.

Now, using the definition of electric potential, we can write V = W/q = - (y)(q1q2)/(2r^2q) + C/q. Since we are interested in the potential caused by a charge Q at (a,b), we can set q1 = Q and r = √[(x-a)^2 + (y-b)^2]. This gives us V(x,y) = - (y)(Qq2)/(2[(x-a)^2 + (y-b)^2]) + C/q.

We can determine the constant C by noting that at infinity, the potential should be zero, so we can write V(infinity) = 0 = - (y)(