A charged cork of mass m suspended by a light | vedclass

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(D) The electric field is $\vec{E} = E_x \hat{i} + E_y \hat{j} = 10^5 \hat{i} + 10^5 \hat{j} \ NC^{-1}$.Let $q$ be the charge on the cork. The forces

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(D) The electric field is $\vec{E} = E_x \hat{i} + E_y \hat{j} = 10^5 \hat{i} + 10^5 \hat{j} \ NC^{-1}$.Let $q$ be the charge on the cork. The forces acting on the cork in equilibrium are:$1$. Tension $T$ in the string at an angle $\alpha$ with the vertical.$2$. Gravitational force $mg$ acting downwards.$3$. Electric force $\vec{F}_e = q\vec{E} = qE_x \hat{i} + qE_y \hat{j}$.Resolving forces in horizontal and vertical directions:Horizontal: $T \sin \alpha = qE_x$Vertical: $T \cos \alpha + qE_y = mg \implies qE_y = mg - T \cos \alpha$Dividing the two equations:$\frac{qE_x}{qE_y} = \frac{T \sin \alpha}{mg - T \cos \alpha}$Since $E_x = E_y$,we have $1 = \frac{T \sin \alpha}{mg - T \cos \alpha}$,which gives $mg - T \cos \alpha = T \sin \alpha$,or $mg = T(\sin \alpha + \cos \alpha)$.Given $T = \frac{2mg}{1 + \sqrt{3}}$,substitute this into the equation:$mg = \frac{2mg}{1 + \sqrt{3}} (\sin \alpha + \cos \alpha)$$1 + \sqrt{3} = 2(\sin \alpha + \cos \alpha)$$\sin \alpha + \cos \alpha = \frac{1 + \sqrt{3}}{2} = \frac{1}{2} + \frac{\sqrt{3}}{2} = \sin 30^o + \cos 30^o$ or $\sin 60^o + \cos 60^o$.Thus,$\alpha = 30^o$ or $60^o$.

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