# The electric field of a charged semicircle from its electric potential

• sergioro
In summary, the electric field due to a uniform charge distribution λ spread out on a semicircle of radius R is computed to be - grad(V)=-d(V(z))/dz.
sergioro

## Homework Statement

A question involving the relationship between the electric
field E(r) and the electric potential V( r) is about computing the electric
field on the z-axis due to a uniform line charge distribution λ
spread out on a semicircle of radius R, lying on the first
half of the XY plane (0 ≤ θ ≤ π). A few points on the
semicircle are (x=R,y=0) at θ= 0, (x=0,y=R) at θ= π/2,
and (x=-R,y=0) at θ = π (I am using the symbol π as Pi).

## Homework Equations

The electric potential V( r) is easily computed giving the result
V(z) = K*Q/Sqrt(z^2 + R^2). Q = λπR and K is the electric constant.

## The Attempt at a Solution

The electric field is suppose to be
obtained via E = - grad(V) = - d(V(z))/dz. But this relation leads only
to one component of the electric field.

Could somebody point out what is missing?

Sergio

Welcome to PF!

Hi Sergio! Welcome to PF!
sergioro said:
The electric field is suppose to be
obtained via E = - grad(V) = - d(V(z))/dz.

No, the gradient of V is the vector (∂V/∂x,∂V/∂y,∂V/∂z)

sergioro said:
The electric potential V( r) is easily computed giving the result
V(z) = K*Q/Sqrt(z^2 + R^2). Q = λπR and K is the electric constant.
Yes, but only knowing the potential on the z axis is not enough information to compute the field there. You also need to know how the potential will change if you move some infinitesimal off that axis. Can you compute the full potential V=V(x,y,z)?

All right, haruspex . I see the point now. Some how I was mislead
by the textbook.

After writing the potential at any point in space (x,y,z) and
Leaving it in integral form, I then computed its derivatives
with respect to each one of the components of the field
point (x,y,z). Then, setting x=y=0 on each one,
integration respect to the source variable yields the expected answer.

Thanks,

Sergio

,

I would respond by saying that the missing component of the electric field can be calculated by considering the symmetry of the problem. Since the semicircle is symmetric about the z-axis, the electric field in the x and y directions will cancel out due to the opposite contributions from the positive and negative charges on either side of the z-axis. Therefore, the only non-zero component of the electric field will be in the z-direction, which can be calculated using the equation E = -d(V(z))/dz as mentioned in the attempt at a solution. This approach takes into account the fact that the electric field is conservative and only depends on the potential function, not on the path taken to reach a certain point. Additionally, the electric field can also be calculated using the formula E = kQz/(z^2 + R^2)^(3/2), where k is the electric constant, Q is the total charge on the semicircle, and z is the distance from the center of the semicircle to the point where the electric field is being calculated. This formula takes into account the contribution from all the individual charges on the semicircle and can be used to verify the result obtained from the potential function.

## 1. What is the formula for calculating the electric field of a charged semicircle?

The formula for calculating the electric field of a charged semicircle is E = (kQ)/(R^2 + x^2)^(3/2), where k is the Coulomb's constant, Q is the charge of the semicircle, R is the radius of the semicircle, and x is the distance from the center of the semicircle to the point where the electric field is being calculated.

## 2. How does the electric field of a charged semicircle vary with distance?

The electric field of a charged semicircle decreases with distance. As the distance from the center of the semicircle increases, the electric field strength decreases.

## 3. Can the electric field of a charged semicircle ever be zero?

Yes, the electric field of a charged semicircle can be zero at the center of the semicircle. This is because at the center, the distance (x) in the formula for electric field becomes zero, making the whole equation zero.

## 4. How does the charge of the semicircle affect its electric field?

The charge of the semicircle directly affects its electric field. As the charge increases, the electric field also increases. This means that a semicircle with a larger charge will have a stronger electric field compared to a semicircle with a smaller charge.

## 5. Can the electric field of a charged semicircle be negative?

Yes, the electric field of a charged semicircle can be negative. This happens when the electric field is directed towards the center of the semicircle, which can occur when the distance (x) in the formula for electric field is greater than the radius (R) of the semicircle.

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