Question

The force between two point charges placed in vacuum at distance 1 mm is 18n. if a glass plate of thickness 1 mm and dielectric constant 6, be kept between the charges then new force between them would be

Answer 1

When dielectric medium is insert new force will be 3 N

Explanation:

We have given force between two charges in vacuum at a distance of 1 mm is 18 N

According to coulomb's law $18=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}$-----eqn 1

Now a glass of dielectric constant 6 is placed between them

So force $F_{new}=\frac{1}{4\pi\times k \epsilon _0}\frac{q_1q_2}{r^2}$

From the relation we can see that new force will be $\frac{1}{k}$ times

So $F_{new}=\frac{18}{6}=3N$

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Answer 2

Answer:

have given force between two charges in vacuum at a distance of 1 mm is 18 N

According to coulomb's law 18=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}18=

4πϵ

0

1

r

2

q

1

q

2

-----eqn 1

Now a glass of dielectric constant 6 is placed between them

So force F_{new}=\frac{1}{4\pi\times k \epsilon _0}\frac{q_1q_2}{r^2}F

new

=

4π×kϵ

0

1

r

2

q

1

q

2

From the relation we can see that new force will be \frac{1}{k}

k

1

times

So F_{new}=\frac{18}{6}=3NF

new

=

6

18

=3N

Learn more

The force between two point charge separated by a distance d in air is 200 N.The force of between the same two charges separated by twice the distance in a medium of dielectric constant 2 is

https://brainly.in/question/1013911