Question 805: Electrostats & Capacitance

Question

A particle ' $A$ ' has charge ' $+q$ ' and a particle ' $B$ ' has charge ' $+4 q$ '. Each has same mass ' $m$ '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)

Accepted Answer

When a charged particle is placed in an electric field, it experiences an electric force. Consider the following for particles A and B:
Electric Force and Acceleration
For particle A:
$    F_A = qE    $
$    F_A = m_A a_A    $
$    	herefore \quad a_A = \frac{qE}{m}    $
For particle B:
$    F_B = 4qE    $
$    F_B = m_B a_B    $
$    a_B = \frac{4qE}{m}    $
From this, it follows that:
$    a_A = \frac{a_B}{4} \quad 	ext{.... (i)}    $
Relation of Speeds
Since both particles start from rest and fall through the same distance, we can use the equations of motion:
For particle A:
$    V_A^2 = 0 + 2a_Ax    $
For particle B:
$    V_B^2 = 0 + 2a_Bx    $
Taking the ratio of their speeds:
$    \left(\frac{V_A}{V_B}ight)^2 = \frac{a_A}{a_B}    $
Using the relationship from (i):
$    \left(\frac{V_A}{V_B}ight)^2 = \frac{a_B}{4 	imes a_B} = \frac{1}{4}    $
Therefore, the ratio of their speeds is:
$    \frac{V_A}{V_B} = \frac{1}{2}    $
In summary, the ratio of the speeds of particles A to B is $1:2$.

Answer

$2: 1$

Answer

$1: 2$

Answer

$1: 4$

Answer

$4: 1$

Electrostats & Capacitance

A particle ' A ' has charge ' +q ' and a particle ' B ' has charge ' +4 q '. Each has same mass ' m '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)

Choose the correct answer: