A ball of mass 2 kg and charge 1 uC is dropped from the top of a tower. In space, the electric...

Question:

A ball of mass 2 kg and charge 1 uC is dropped from the top of a tower. In space, the electric field exists in the horizontal direction away from the tower which varies as E = (5 - 2x) {eq}\times 10^6 {/eq}. Find the maximum horizontal distance the ball can go.

Electric force:

When two charged particle lies in the same plane, they exert force on each other that can be attractive or repulsive in nature; this force is termed as an electric force. The electric force depends on the magnitude of the charge on the particle and nature of the charge.

Answer and Explanation:

Given data

  • Mass of ball {eq}m = 2\,{\rm{kg}} {/eq}
  • Charge on ball {eq}q = 1\;{\rm{\mu C}} {/eq}
  • Electric field {eq}E = \left( {5 - 2x} \right) \times {10^6}\;{\rm{V/m}} {/eq}
  • The expression of acceleration due to electric force,

{eq}a = \dfrac{{qE}}{m}...........................\left( 1 \right) {/eq}


The expression of acceleration in linear motion,

{eq}\begin{align*} a &= \dfrac{{dv}}{{dt}}\\ a &= \dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}}\\ a &= v\dfrac{{dv}}{{dx}}.....................\left( 2 \right) \end{align*} {/eq}

Here, v is the velocity, t is the time and x id the distance.


From equation (1) and (2),

{eq}\dfrac{{qE}}{m} = v\dfrac{{dv}}{{dx}} {/eq}

Substitute the value of E,

{eq}v\dfrac{{dv}}{{dx}} = \dfrac{q}{m}\left( {5 - 2x} \right) \times {10^6}........................\left( 3 \right) {/eq}


Integrate equation (3),

{eq}\begin{align*} \int {vdx} &= \dfrac{q}{m} \times {{10}^6\int {\left( {5 - 2x} \right)} } dx\\ \dfrac{{{v^2}}}{2} &= \dfrac{q}{m}\left( {5x - {x^2}} \right)\\ {v^2} &= \dfrac{{2q}}{m}\left( {5x - {x^2}} \right)\\ v &= \sqrt {\dfrac{{2q}}{m}\left( {5x - {x^2}} \right)} ...................\left( 4 \right) \end{align*} {/eq}


The expression for velocity in linear motion,

{eq}v = \dfrac{{dv}}{{dx}}...................................\left( 5 \right) {/eq}


From equation (4) and (5),

{eq}\dfrac{{dx}}{{dt}} = \sqrt {\dfrac{{2q}}{m} \times {{10}^6}\left( {5x - {x^2}} \right)} ..............\left( 6 \right) {/eq}


For x to be maximum,

{eq}\dfrac{{dx}}{{dt}} = 0 {/eq}

Substitutes the value in equation (6),

{eq}\begin{align*} \sqrt {\dfrac{{2q}}{m} \times {{10}^6}\left( {5x - {x^2}} \right)} & = 0\\ 5x - {x^2}& = 0\\ x &= 5\;{\rm{m}} \end{align*} {/eq}

Thus, maximum horizontal distance is 5 m.


Learn more about this topic:

Loading...
Electric Fields Practice Problems

from Physics 101: Intro to Physics

Chapter 17 / Lesson 8
3.4K

Related to this Question

Explore our homework questions and answers library