A thin nonconducting rod is bent to form a semi-circle of radius R. A charge Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown int he sketch at the right.find E vector at P, the center of the semi-circle. the semi circle looks like that, plus at the top and minuses at the bottom. P is the center and the radius arrow point to the minuses

Question

A thin nonconducting rod is bent to form a semi-circle of radius R. A charge Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown int he sketch at the right.find E vector at P, the center of the semi-circle. the semi circle looks like that, plus at the top and minuses at the bottom. P is the center and the radius arrow point to the minuses

Answer 1

So I assume you have calculus based physics.

integrate the charge through and angle. let theta begin at the bottom rotate 90 deg for all the neg charge, then from 90 to 180 for the positive charge.

E=INT kdq/r^2 where dq=Q/(pi*r/2)dTheta

integrate from 0 to PI/2, that is E for the lower charge, and it points to the center point of the quadrant (45 deg, by symettry).
Do the same for the + charge, it points away from the center of the upper quadrant.

Now add the two E components. Neat, it point downward, and has value of E*sqrt2
work that out.

Answer 2

To find the electric field, E, at point P, the center of the semi-circle, due to the charges distributed along the semi-circle, we can divide the problem into smaller parts and use the principle of superposition.

Let's consider one small segment of the semi-circle (dθ) with a small charge (dQ) on it. The electric field, dE, at point P due to this small charge can be calculated using Coulomb's Law:

dE = (1/4πε₀) * (dQ/r²) * r * dθ

Where ε₀ is the vacuum permittivity, r is the distance from the small charge to point P, and dθ is the small angle subtended by the small segment.

Since the charge distribution is uniform, we can consider the entire semi-circle as a collection of infinitesimally small charge segments, each producing a small electric field. To calculate the electric field at point P, we need to integrate the contributions from all these small charge segments.

Before we proceed with the integration, let's consider the symmetries of the problem. Since there is no component of the electric field in the vertical direction (perpendicular to the plane of the semi-circle), the only non-zero component of the electric field will be in the horizontal direction (tangent to the semi-circle). Therefore, we only need to calculate the x-component of the electric field, Ex, at point P.

Now, let's do the integration. Since the charge is uniformly distributed, the total charge, Q, is divided equally among all the small charge segments. Therefore, each small charge segment has a charge of dQ = Q/(2πR) * R * dθ.

Substituting this into the expression for dE, we have:

dE = (1/4πε₀) * (Q/(2πR) * R * dθ / r²) * r * dθ

Simplifying, we get:

dE = (1/8πε₀) * (Q/R) * sin(θ) dθ

Now, we can integrate this expression from 0 to π radians (the half-circle) to get the total x-component of the electric field at point P:

E = ∫(0 to π) dE = ∫(0 to π) (1/8πε₀) * (Q/R) * sin(θ) dθ

Evaluating this integral gives us the x-component of the electric field at point P, as it is indicated by the arrow pointing towards the negative charges at the bottom of the semi-circle.