A block of mass m containing a net negative charge -q is placed on a frictionless horizontal t

English

Gujarati

Hindi

2. Electric Potential and Capacitance

medium

A block of mass $m$ containing a net negative charge $-q$ is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant $k$ as shown. If horizontal electric field $E$ parallel to the spring is switched on, then the maximum compression of the spring is :-

A

$ \sqrt {qE / k}$

B

$ \sqrt {k / qE}$

C

$qE/k$

D

$\frac{2qE}{k}$

Solution

At the point of max compression

Work done by spring $=$ Potential due to the charge

$W_{\text {spring}}=V_{q}$

$\frac{1}{2} k x^{2}=E q$

$x=\sqrt{\frac{2 E q}{k}}$

Standard 12

Physics

Share

Similar Questions

A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is

hard

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

medium

The work done in moving an electric charge $q$ in an electric field does not depend upon

easy

Two identical thin rings, each of radius $R $ meter are coaxially placed at distance $R$ meter apart. If $Q_1$ and $Q_2$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is

hard

If an electron moves from rest from a point at which potential is $50\, volt$ to another point at which potential is $70\, volt$, then its kinetic energy in the final state will be

easy