Question

A thin glass rod is bent into a semicircle 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 in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

Short answer

Answer: The electric field at the center of the semicircle is 0.

Step-by-step solution

Find the charge per unit length

We are given that the total charge for both the positive and negative charges is \(Q\). Since the charges are uniformly distributed, we can find the charge per unit length by dividing the total charge by length. The circumference of the semicircle is given by \(L=\pi R\). So, the charge per unit length is: $$\lambda_+ = \frac{Q}{\pi R}$$ for the positive charges and $$\lambda_- = -\frac{Q}{\pi R}$$ for the negative charges.

Find the electric field for a small charge element

Consider a small length \(d\theta\) of the semicircle at an angle \(\theta\) with the positive x-axis as shown in the figure. The length of this small arc is given by \(Rd\theta\). The charge on this small arc is given by \(dq = \lambda_+ Rd\theta\) for the positive charges and \(dq = \lambda_- Rd\theta\) for the negative charges. The electric field produced by this small charge at point P is given by the formula for electric field due to a point charge: $$d\vec{E} = \frac{k dq}{R^2}$$ where \(k\) is Coulomb's constant.

Break down electric field vectors into components

The electric field produced by the small charge element lies in the plane formed by the semicircle and has x and y components. The x and y components of the electric field are given by: $$dE_x = dE \cos\theta$$ $$dE_y = dE \sin\theta$$

Integrate electric field components over the semicircle

To find the total electric field at point P, we need to integrate the electric field components over the entire semicircle. Integrate each component separately for the positive and negative charges and add them up. The limits of integration are from \(0\) to \(\pi\). For the positive charges: $$E_{x+} = \int_0^\pi \frac{k\lambda_+ R d\theta}{R^2} \cos\theta = k\lambda_+ \int_0^\pi \frac{d\theta}{R} \cos\theta$$ $$E_{y+} = \int_0^\pi \frac{k\lambda_+ R d\theta}{R^2} \sin\theta = k\lambda_+ \int_0^\pi \frac{d\theta}{R} \sin\theta$$ For the negative charges: $$E_{x-} = \int_0^\pi \frac{k\lambda_- R d\theta}{R^2} \cos\theta = k\lambda_- \int_0^\pi \frac{d\theta}{R} \cos\theta$$ $$E_{y-} = \int_0^\pi \frac{k\lambda_- R d\theta}{R^2} \sin\theta = k\lambda_- \int_0^\pi \frac{d\theta}{R} \sin\theta$$

Calculate the total electric field components

Add the electric field components of the positive and negative charges to get the total electric field components at point P: $$E_x = E_{x+} + E_{x-} = k\lambda_+ \int_0^\pi \frac{d\theta}{R} \cos\theta + k\lambda_- \int_0^\pi \frac{d\theta}{R} \cos\theta$$ $$E_y = E_{y+} + E_{y-} = k\lambda_+ \int_0^\pi \frac{d\theta}{R} \sin\theta + k\lambda_- \int_0^\pi \frac{d\theta}{R} \sin\theta$$ Solving the above integrals, we get: $$E_x = 0$$ $$E_y = 0$$

Find the magnitude and direction of the electric field

Since the electric field components are zero, the magnitude of the electric field at point P is also zero. Therefore, there is no need to find the direction of the electric field as it doesn't exist. So, the electric field \(\vec{E}\) at the center of the semicircle is 0.