Question

Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small angle Δθ=Δs /R, where R is the radius. Express your answer in terms of the variables Q, L, unit vectors i^, j^, and appropriate constants.

Answer 1

Answer:

Electric Field a the centre$E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)$

Explanation:

Given:

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle $\theta$ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let $\lambda$ be the charge per unit length such that$\lambda=\dfrac{Q}{\pi R}$

$k=\dfrac{1}{4\pi \epsilon_0}$

Total Electric Field at the centre

$=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta$

integrating 0 to $\dfrac{\pi}{2}$

$E=\dfrac{2k\lambda}{R}(-\vec j)$

$E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)$

So the Electric field at the centre is calculated.