In a region,the electric field varies as E = | vedclass

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

Question

(C) The electric field is given by $E = 2x^2 - 4$. For the charge to cross the origin from infinity with minimum velocity,it must just reach the point

Vedclass · Accepted Answer

The kinetic energy at $x = \sqrt{2} \, m$ must be zero.

Vedclass · Answer

(C) The electric field is given by $E = 2x^2 - 4$. For the charge to cross the origin from infinity with minimum velocity,it must just reach the point where the electric field becomes zero,as the field opposes the motion.Setting $E = 0$:$2x^2 - 4 = 0$$x^2 = 2$$x = \sqrt{2} \, m$ (considering the region of interest).At $x = \sqrt{2} \, m$,the electric field changes sign. Since the charge is released from infinity (where potential is zero) and moves towards the origin,it experiences a retarding force until it reaches $x = \sqrt{2} \, m$. For the minimum velocity condition,the kinetic energy must be zero at the point where the electric field is zero,which is $x = \sqrt{2} \, m$.

In a region,the electric field varies as $E = 2x^2 - 4$,where $x$ is the distance in $m$ from the origin along the $x$-axis. $A$ positive charge of $1 \,\mu C$ is released with minimum velocity from infinity to cross the origin. Then:

Similar Questions

If the potential at the centre of a uniformly charged hollow sphere of radius $R$ is $V$,then the electric field at a distance $r$ from the centre of the sphere is $(r > R)$.

What is the magnitude of a point charge which produces an electric field of $2 \, N/C$ at a distance of $60 \, cm$? (Given: $1/(4\pi \varepsilon_0) = 9 \times 10^9 \, N \cdot m^2/C^2$)

Three charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the centre of the triangle:

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of the electric field at point $C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module