Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

1 answer:

3 0

We have that for the Question "Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

$Q=\frac{E}{Nr^2}$

From the question we are told

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the magnitude of charge moved</h3>

Generally the equation for the magnitude of charge moved, Q is mathematically given as

$Q=\frac{E}{Nr^2}$

Therefore

An expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e" it can be

$Q=\frac{E}{Nr^2}$

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Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

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$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$