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(C) Given,linear charge density $\lambda = \lambda_0 \cos \phi$. Consider two symmetric elements of length $dl = r d\phi$ at an angle $\phi$ on both s

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$\frac{\lambda_0}{4 \varepsilon_0 r}$

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(C) Given,linear charge density $\lambda = \lambda_0 \cos \phi$. Consider two symmetric elements of length $dl = r d\phi$ at an angle $\phi$ on both sides of the vertical axis. The charge on each element is $dq = \lambda dl = (\lambda_0 \cos \phi) r d\phi$. The electric field due to each element at the centre is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} = \frac{\lambda_0 \cos \phi d\phi}{4 \pi \varepsilon_0 r}$. The horizontal components $dE \sin \phi$ cancel out due to symmetry. The net electric field is the sum of vertical components $dE \cos \phi$: $E = \int_0^{\pi} 2 (dE \cos \phi) = \int_0^{\pi} 2 \left( \frac{\lambda_0 \cos \phi d\phi}{4 \pi \varepsilon_0 r} \right) \cos \phi$ $E = \frac{2 \lambda_0}{4 \pi \varepsilon_0 r} \int_0^{\pi} \cos^2 \phi d\phi$ Using $\cos^2 \phi = \frac{1 + \cos 2\phi}{2}$: $E = \frac{\lambda_0}{4 \pi \varepsilon_0 r} \int_0^{\pi} (1 + \cos 2\phi) d\phi = \frac{\lambda_0}{4 \pi \varepsilon_0 r} [\phi + \frac{\sin 2\phi}{2}]_0^{\pi} = \frac{\lambda_0}{4 \pi \varepsilon_0 r} [\pi - 0] = \frac{\lambda_0}{4 \varepsilon_0 r}$.

$A$ thin non-conducting ring of radius $r$ has a linear charge density $\lambda = \lambda_0 \cos \phi$,where $\lambda_0$ is a constant and $\phi$ is the azimuthal angle. The magnitude of the electric field strength at the centre of the ring is

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