Question

A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged uniformly with a total charge Q, what is the electric field at P? (Assume Q is positive. For the magnitude, use the following as necessary: ε0, Q, R, and θ.)

Answer 1

To find the electric field at point P, we can integrate the electric field contribution from each small segment of the bent rod. The electric field due to a small segment of length Rdθ is given by dE = (kQ/4πε₀R) * cosθ * dθ. By integrating this expression over the range of angles from -θ to θ, we can find the total electric field at P.

Step-by-step explanation:

The electric field at point P can be found by applying the superposition principle to symmetrically placed charge elements and integrating. For a rod bent into the arc of a circle subtending an angle of 2θ at the center P, the electric field at P due to a small segment of the rod is given by dE = (kQ/4πε₀R) * cosθ * dθ. By integrating this expression over the range of angles from -θ to θ, we can find the total electric field at P.

Answer 2

Qsinθ/4πε₀R²θ

Step-by-step explanation:

Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.

Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.

So dq = λRdθ.

Substituting dq into dE', we have

dE' = dqcosθ/4πε₀R²

= λRdθcosθ/4πε₀R²

= λdθcosθ/4πε₀R

E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ

E' = (λ/4πε₀R)[sinθ] from -θ to θ

E' = (λ/4πε₀R)[sinθ]

= (λ/4πε₀R)[sinθ - sin(-θ)]

= (λ/4πε₀R)[sinθ + sinθ]

= 2(λ/4πε₀R)sinθ

= (λ/2πε₀R)sinθ

Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ

Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ

Q = 2λRθ

λ = Q/2Rθ

Substituting λ into E', we have

E' = (Q/2Rθ/2πε₀R)sinθ

E' = (Q/θ4πε₀R²)sinθ

E' = Qsinθ/4πε₀R²θ where θ is in radians