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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$A$ and $B$ both

Vedclass · Answer

(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 $60^o$.

Similar Questions

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.

In the following four situations,charged particles are at an equal distance from the origin. Arrange them in order of the magnitude of the net electric field at the origin,starting from the greatest.

The Van de Graaff generator is not based on

$A$ particle of mass $m$ and charge $q$ is thrown in a region where uniform gravitational field and electric field are present. The path of the particle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module