Question

A particle of mass m and positive charge q is released from point A. It's speed is found to be v, when it passed through a point B. Which of the 2 points is at higher potential? What is the potential difference between the points?

Answer 1

Given,

A particle of mass m and positive charge q is released from point A. It's speed is found to be v, when it passed through a point B.

To find,

which of the 2 points is at higher potential ?

what is the potential difference between the points ?

concept : a positive charge always moves from higher potential to lower potential.

so if a positive charge particle moves from point A to point B, then definitely point A is at higher potential.

from work energy theorem,

workdone by electric field = change in kinetic energy of particle

⇒q ∆V = 1/2 mv² - 1/2 m (0)² = 1/2 mv²

⇒∆V = mv²/2q

Therefore potential difference between the points is mv²/2q.

Answer 2

$\huge{\underline{\bf\red{QuestiØn} :}}$

$\sf A \: particle, \: of \: mass \: m \: and \: positive \: charge \: q, \: \\ \sf is \: released \: from \: point \: A. \: It's \: speed \: is \: found \: to \: be \: v, \\ \sf when \: it \: passed \: through \: a \: point \: B. \\ \sf \: \bigstar \: Which \: of \: the \: 2 \: points \: is \: at \: higher \: potential \:? \qquad \qquad \\ \sf \bigstar \: \: What \: is \: the \: potential \: difference \: between \: the \: points \: ?$

$\huge{\underline{\bf\blue{SolutiØn}\ :}}$

Given :-

• mass of the particle = m

• charge = ( + ) q [ when released from A ]

• speed = v [ when passed through B ]

We know that ,

A positive ( + ) charge always moves from higher potential to lower potential.

Thus ,

$\bigstar$If a positively charged particle moves from point A to point B , then it's obvious that — point A is at higher potential than that of B .

Now,

we know that,

A/c to WORK ENERGY THEORUM,

$\begin{aligned} \\ \sf work \: done \: by& = \sf change \: in \: kinetic \\ \sf electric \: field& \quad \ \ \sf energy \: of \: particle \\ \\ \implies \qquad \quad \sf q \Delta V & = \sf \frac{1}{2}m {v}^{2} - \frac{1}{2} {(0)}^{2} \\ \\ & = \sf \frac{1}{2} m {v}^{2} \\ \\ \implies \sf \quad \qquad \ \ \Delta V & = \sf \frac{m {v}^{2} }{2q} & & & & & & \end{aligned}$

$\bigstar$$\mathbb{ HENCE, } \: \sf the \: potential \: difference \: between \: the \: points \: is \: \boxed{ \sf \frac{m {v}^{2} }{2q} }.$

$\red{\rule{5.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$