Solveeit Logo
Login

Question

Physics Question on electrostatic potential and capacitance

Electric field at centre O of semicircle of radius a having linear charge density λ\lambda given as

A

2λε0a \frac{ 2 \lambda}{ \varepsilon_0 a}

B

λπε0a \frac{ \lambda \pi}{ \varepsilon_0 a}

C

λ2πε0a \frac{ \lambda}{2 \pi \varepsilon_0 a}

D

λπε0a \frac{ \lambda}{ \pi \varepsilon_0 a}

Answer

λ2πε0a \frac{ \lambda}{2 \pi \varepsilon_0 a}

Explanation

Solution

Considering symmetric elements each of length dldl at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length dl(=adθ)dl(=ad\theta ) along PO.
dE=14πε0dqa2cos⁡θdE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{dq}{{{a}^{2}}}\cos \theta
(∵dl=adθ)(\because dl=ad\theta )
=14πε0λdla2cos⁡θ=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dl}{{{a}^{2}}}\cos \theta
=14πε0λ(adθ)a2cos⁡θ=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda (ad\theta )}{{{a}^{2}}}\cos \theta Net electric field at O
E=∫−π/2π/2dE=2∫Oπ/214πε0λacos⁡θ dθa2E=\int_{-\pi /2}^{\pi /2}{dE=2\int_{O}^{\pi /2}{\frac{1}{4\pi {{\varepsilon }_{0}}}}}\frac{\lambda a\cos \theta \,d\theta }{{{a}^{2}}}
=2⋅14πεOλa[sin⁡θ]oπ/2=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\frac{\lambda }{a}[\sin \theta ]_{o}^{\pi /2}
=2⋅14πεO⋅λa⋅1=λ2πεoa=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {{\varepsilon }_{o}}a}