A thin glass rod is bent in a semicircle of r | vedclass

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Question

(D) The linear charge density is given by $\lambda = \lambda_0 \sin 	heta$. Consider a small element of the rod subtending an angle $d	heta$ at the

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The linear charge density is given by $\lambda = \lambda_0 \sin 	heta$. Consider a small element of the rod subtending an angle $d	heta$ at the centre. The charge on this element is $dq = \lambda (R d	heta) = \lambda_0 R \sin 	heta d	heta$.The electric field $dE$ at the centre due to this element is $dE = \frac{k dq}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda_0 R \sin 	heta d	heta}{R^2} = \frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin 	heta d	heta$.Due to symmetry, the $x$-components of the electric field cancel out. The $y$-component is $dE_y = -dE \sin 	heta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin^2 	heta d	heta$.Integrating from $	heta = 0$ to $\pi$:$E_y = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \sin^2 	heta d	heta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \frac{1 - \cos 2	heta}{2} d	heta = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} [	heta - \frac{\sin 2	heta}{2}]_0^{\pi} = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} (\pi) = -\frac{\lambda_0}{8 \varepsilon_0 R}$.Thus, $\vec{E} = -\frac{\lambda_0}{8 \varepsilon_0 R} \hat{j}$. Since this is not among the options, the correct answer is $D$.

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