KCET 2018 Physics Solved Paper

Given, force of repulsion between two identical positive charges with separation, r = F.
Half of gap is filled with dielectric constant = 4.
Now,F=‌

1

4πε0

‌

q1q2

r2

where q1 and q2 are two positive charges. For identical charges, q1=q−2=q
F=‌

1

4πε0

‌

q2

r2

→ (1)
When the gap is half filled with dielectric constant then
F′=‌

1

4πε0

‌

q2

(r−t+t√k)2

where k is dielectric constant = 4; t is thickness of dielectric = r/2. Therefore,
Thus F′=

4

9

F   (Using Eq. (1))