Find the force experienced by the semicircula | vedclass

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(B) The electric field $E$ at a distance $R$ from an infinite line charge with linear charge density $\lambda$ is given by $E = \frac{\lambda}{2\pi\va

Vedclass · Accepted Answer

$\frac{{\lambda q}}{{{\pi ^2}{\varepsilon _0}R}}$

Vedclass · Answer

(B) The electric field $E$ at a distance $R$ from an infinite line charge with linear charge density $\lambda$ is given by $E = \frac{\lambda}{2\pi\varepsilon_0 R}$.Consider a small element of the semicircular rod of angular width $d	heta$ at an angle $	heta$ from the perpendicular bisector. The charge on this element is $dq = \left(\frac{q}{\pi R}ight) R d	heta = \frac{q}{\pi} d	heta$.The force on this element is $dF = (dq)E = \left(\frac{q}{\pi} d	hetaight) \left(\frac{\lambda}{2\pi\varepsilon_0 R}ight) = \frac{\lambda q}{2\pi^2\varepsilon_0 R} d	heta$.Due to symmetry,the components of the force perpendicular to the axis of symmetry cancel out. The net force is the sum of the components along the axis of symmetry: $F_{	ext{net}} = \int_{-\pi/2}^{\pi/2} dF \cos	heta$.$F_{	ext{net}} = \int_{-\pi/2}^{\pi/2} \frac{\lambda q}{2\pi^2\varepsilon_0 R} \cos	heta d	heta = \frac{\lambda q}{2\pi^2\varepsilon_0 R} [\sin	heta]_{-\pi/2}^{\pi/2}$.$F_{	ext{net}} = \frac{\lambda q}{2\pi^2\varepsilon_0 R} [1 - (-1)] = \frac{\lambda q}{2\pi^2\varepsilon_0 R} (2) = \frac{\lambda q}{\pi^2\varepsilon_0 R}$.

Find the force experienced by the semicircular rod charged with a charge $q$,placed as shown in the figure. The radius of the wire is $R$ and the line of charge with linear charge density $\lambda$ is passing through its center and is perpendicular to the plane of the wire.

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