Question

14. The ratio of the forces between two
charges placed at a certain distance apart
in air and at half of the distance apart in
medium of dielectric 'k' is
1) 1 : 4k 2) K:4 3) 4k: 1 4) 4:k​

Answer 1

Answer:

Given:

Initial in air , the forces between 2 charge is in air.

In medium , they are located at half distance.

To find:

Ratio of forces

Calculation:

Let initial force be F1 in air.

$F1 = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{(q1)(q2)}{ {d}^{2} }$

Let the force in medium be F2

$F2 = \dfrac{1}{4\pi \epsilon_{0}k} \dfrac{(q1)(q2)}{ {( \frac{d}{2} )}^{2} }$

$= > F2 = \dfrac{1}{4\pi \epsilon_{0}k} \dfrac{4(q1)(q2)}{ {d}^{2} }$

Now , required ratio :

$\boxed{ \red{ \huge{ \sf{F1 : F2 = k : 4}}}}$

Answer 2

Answer:

Answer:

Given:

Initial in air , the forces between 2 charge is in air.

In medium , they are located at half distance.

To find:

Ratio of forces

Calculation:

Let initial force be F1 in air.

F1 = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{(q1)(q2)}{ {d}^{2} }F1=

4πϵ

0

1

d

2

(q1)(q2)

Let the force in medium be F2

F2 = \dfrac{1}{4\pi \epsilon_{0}k} \dfrac{(q1)(q2)}{ {( \frac{d}{2} )}^{2} }F2=

4πϵ

0

k

1

(

2

d

)

2

(q1)(q2)

= > F2 = \dfrac{1}{4\pi \epsilon_{0}k} \dfrac{4(q1)(q2)}{ {d}^{2} }=>F2=

4πϵ

0

k

1

d

2

4(q1)(q2)

Now , required ratio :

\boxed{ \red{ \huge{ \sf{F1 : F2 = k : 4}}}}

F1:F2=k:4