# Electric Fields-Calculating the total force on a charge

• rico22
In summary: You know what the limits of integration are.In summary, the problem involves a semicircle of positive charge with a charge per unit length described by λ = λ0 cos(θ). The total charge on the semicircle is given as 15.0 µC. The magnitude of the total force on a charge of 2.00 µC placed at the center of curvature is to be calculated. The electric field is found by integrating the expression kλ/R∫sin(θ) and multiplying it by the charge of 2.00µC. The direction of the field is downward (negative y) and the limits of integration are from -pi/2 to +pi/2.
rico22

## Homework Statement

A line of positive charge is formed into a semicircle of radius R = 40.0 cm. The charge per unit length along the semicircle is described by the expression λ = λ0 cos(θ). The total charge on the semicircle is 15.0 µC. The semi circle starts off at 0 and finishes at pi going through quadrants I and II. Calculate the magnitude of the total force on a charge of 2.00 µC placed at the center of curvature. θ is to be taken to start off from pi/2 thus creating a triangle with R that cos(θ)= y/R and sin(θ)=x/R

λ=Q/l
l=pi/R
∫dE=∫(k ldql/r2

## The Attempt at a Solution

I replaced ldql with λRdθ. then I found lambda to be 1.194x10-5 since I know the length of the semicircle and total charge Q. I plugged this into the integral to find E(electric field) and got Ex=kλ/R∫sin(θ) ⇔ kλ/R (-cosθ). this should give me the magnitude of the electric field. then I multiplied this time the charge of 2.00μC placed at the center of the curvature. Where am I going wrong? Any response would be greatly appreciated.

Don't forget that $\lambda$ is a function of $\vartheta$: $\lambda(\vartheta)$.

so since I have Q can I just integrate dq=λds=λrdθ?

so Q=∫rλdθ=r∫λdθ? but this is where I get confused should I integrate λcos(θ)?

It would probably be easiest to do each component separately.

you mean r[∫λ∫cos(θ)]?

I mean am I even on the right path with my first attempt? Is this the only thing I did wrong?

anybody?

rico22 said:
so since I have Q can I just integrate dq=λds=λrdθ?

so Q=∫rλdθ=r∫λdθ? but this is where I get confused should I integrate λcos(θ)?
Yes. That's how to find λ0.
I found the statement of the problem a bit confusing, but I think it is measuring theta as the angle from the positive Y axis, clockwise being positive for the angle. So theta runs from -pi/2 at (-R, 0) through 0 at (0, R) to pi/2 at (R, 0). Is that your understanding?
On that basis, can you immediately deduce what direction the field must be in at the origin?

thank you for the reply...yes I know that the limits of integration would be from -pi/2 to +pi/2... I am just confused as to whether I am in the right track as far as the integration goes being that λ in this problem is a function of θ.

as far as the direction of the field goes I would guess that its in the downward (negative y) direction?

rico22 said:
as far as the direction of the field goes I would guess that its in the downward (negative y) direction?
Yes, which means you only need to include the component in that direction in the integrand.
But first, do the integration of the charge to find λ0.

But first, do the integration of the charge to find λ0

Well that's what I am having problems with being that λ is a function of θ, usually I would just take it out of the integrand...so in this case ∫λ0cos(θ) would not equal λ0sin(θ) from -pi/2 to +pi/2? correct?

rico22 said:
so in this case ∫λ0cos(θ) would not equal λ0sin(θ) from -pi/2 to +pi/2? correct?

Why not?

## 1. What is an electric field?

An electric field is a physical quantity that describes the influence of electric charges on other charges in the surrounding space. It is a vector field, meaning it has both magnitude and direction.

## 2. How is the electric field calculated?

The electric field can be calculated by dividing the force acting on a charged particle by the magnitude of the charge. This can be represented by the equation E = F/q, where E is the electric field, F is the force, and q is the magnitude of the charge.

## 3. What is the unit of measurement for electric field?

The unit of measurement for electric field is newtons per coulomb (N/C), which represents the amount of force exerted on a charged particle per unit of charge.

## 4. How do multiple charges affect the electric field?

If there are multiple charges present, the electric field can be calculated by taking the vector sum of the individual electric fields at a particular point. This means that the total electric field is the sum of the individual electric fields created by each charge.

## 5. What is the relationship between electric field and electric force?

The electric force acting on a charged particle is directly proportional to the electric field at that point. This means that the stronger the electric field, the greater the force on the charged particle will be.

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