Question 567: Electrostats & Capacitance

Question

In free space, a particle A of charge 1$$\mu $$C is held fixed at a point P. Another particle B of the same charge and
mass 4$$\mu $$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P
is :
$$\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 	imes {{10}^9}N{m^2}{C^{ - 2}}} ight]$$

Accepted Answer

q
A
= 1 $$\mu $$c ; q
B
= 1 $$\mu $$c, m
B
= 4 ÃÂ 10
Ã¢ÂÂ9
kg, r
AB
= 10
Ã¢ÂÂ3
m
$${1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 	imes {{10}^{ - 3}}}}} ight\}$$
$${1 \over 2}4 	imes {10^{ - 9}}{V^2} = 9 	imes {10^9} 	imes {10^{ - 6}} 	imes {8 \over 9} 	imes {10^3}$$
$${V^2} = {8 \over 2} 	imes {10^9} = 4 	imes {10^9}$$
V =  $$\sqrt {40}  	imes {10^4}\,m/s$$ = 6.32 $$ 	imes $$ 10
4
m/s

Answer

1.0 m/s

Answer

6.32 $$ 	imes $$ 104 m/s

Answer

2.0 $$ 	imes $$ 103 m/s

Answer

1.5 $$ 	imes $$ 102 m/s

Electrostats & Capacitance

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