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\ .