Question

Two small identical fierce x and y have charges + 12 microcoulomb and + 6 microcoulomb. If their separation is 0.5 metre, calculate the force between them when they are immersed in water (K=80)

Answer 1

$\boxed{\textbf{\large{Correct question}}}$

Two small identical spheres x and y have charges + 12 microcoulomb and + 6 microcoulomb. If their separation is 0.5 metre, calculate the force between them when they are immersed in water (K=80)

$\boxed{\textbf{\large{Step by step explanation}}}$

◾As we know the concept of coulomb's law,

Coulomb's Law

The net electric force between two charges is directly proportional to the product of magnitude of two charges and inversely proportional to the square of the distance between them .

◾let us consider, the two charges are Q1 and Q2 and the distance between them is r

so from the Above concept

F α ( Q1 x Q2 ) And F α ( 1 / r^2 )

F =( 1 / 4π€0 ) x (( Q 1 x Q 2) /r^2 )

◾Here, (1 / 4π€0 ) is a permeability constant, where as we have given the medium is water, so for water ( 1 / 4π€0 ) = 80

◾In the given question, we have given Two small identical spheres x and y, having charges +12microcoulomb and + 6 microcoulomb And the distance between them is i.e r = 0.5 m

To convert microcoulomb into coulomb multiply the given value by 10^ ( -6)

therefor 12 microcoulomb = 12 x 10^(-6) And 6 microcoulomb = 6 x 10 ^( - 6 )

Therefor ,

Electric Force

=( 1 / 4π€0 )x (( Q1 x Q2 \) /r^2 )

= ( 80 ) x [( (12 x ( 10^(-6)) x ( (6) x (10 ^ (-6 )) / ( 0.5)^2 ]

= [ ( 80 x 72 x (10 ^ (- 12 )) / 0.25 ]

= [ ( 80 x 72 x ( 10 ^( -12 )) / ( 25 x 10^(-2 ))]

= [( ( 2.88 x 80 ) x (10 ^(-12))/ ( 10 )^( -2 ) ]

= [ 230.4 x ( 10^ ( -12 )) (( 10 ) ^ ( 2 ) )]

= [230.4 x 10^ (- 10) ]

= 23.04 x (10 ) ( 10 ^ ( -10 ))

= 23.04 x ( 10)^(-9)N [ N stands for unit Newton ]

Electric Force

= 23.04 x ( 10)^(-9))

Answer 2

Answer:

hope this will help you.