Two like charges when placed in a medium of dielectric constant 2.5 at a distance of 10 cm repel each
other with a force of 1 N. What will be the force between them when placed 5 cm apart in air ?
Answers
Answer:
Given
two like charges when placed in a medium of dielectric constant of 2.5 at a distance of 0.1 m repel each other with a force of 1 N.
To Find
The force between them when they are placed 0.05 m apart in air.
Working formula
Force between charge particles placed in air
\dfrac{Kq_1q_2}{r^2}
r
2
Kq
1
q
2
Where, K = 9 \times 10^99×10
9
and r is the distance between the charged particles.
Force between charge particles placed in a dielectric medium
Here, if the distance between the particles is r, we use the same formula as that of air ,but the new distance is given as \sqrt{E_or}\:,E_o
E
o
r
,E
o
is the dielectric constant of the medium.
Calculations
0.1 m distance in a dielectric medium of constant 2.5 gives a distance \sqrt{2.5 \times 0.1}\:=\:0.5
2.5×0.1
=0.5 m distance in air.
Hence,when placed 0.5 m apart , force = 1 N. Thus, we have to find force when they are placed 0.05 m apart.
\begin{gathered} \dfrac{F_1}{F_2}\:=\: {(\dfrac{r_2}{r_1})}^2 \\ \\ \\ = > \: \dfrac{1}{F_2}\:=\: {(\dfrac{0.05}{0.5})}^2 \\ \\ \\= > \: \dfrac{1}{F_2} \:=\:\dfrac{1}{100} \\ \\ \\ = > \: F_2\:=\:100\:N \end{gathered}
F
2
F
1
=(
r
1
r
2
)
2
=>
F
2
1
=(
0.5
0.05
)
2
=>
F
2
1
=
100
1
=>F
2
=100N