Find the electric field at point P (as shown | vedclass

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Question

(C) The electric field $E$ at a distance $a$ from the center of a uniformly charged rod of length $L$ and total charge $Q$ on its perpendicular bisect

Vedclass · Accepted Answer

$\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L ^{2}}$

Vedclass · Answer

(C) The electric field $E$ at a distance $a$ from the center of a uniformly charged rod of length $L$ and total charge $Q$ on its perpendicular bisector is given by:$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{a \sqrt{a^2 + (L/2)^2}}$Given $a = \frac{\sqrt{3}}{2} L$,we substitute this into the formula:$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(\frac{\sqrt{3}}{2} L) \sqrt{(\frac{\sqrt{3}}{2} L)^2 + (L/2)^2}}$$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(\frac{\sqrt{3}}{2} L) \sqrt{\frac{3}{4} L^2 + \frac{1}{4} L^2}}$$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(\frac{\sqrt{3}}{2} L) \sqrt{L^2}}$$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(\frac{\sqrt{3}}{2} L) \cdot L} = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\frac{\sqrt{3}}{2} L^2}$$E = \frac{2 Q}{4 \sqrt{3} \pi \varepsilon_{0} L^2} = \frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L^2}$

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a = \frac{\sqrt{3}}{2} L$.

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