JEE Main 25 July 2021 Shift 2 Solved Paper

Section: Physics

Question : 10 of 90

Marks: +1, -0

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb can be expressed as :

[25 Jul 2021 Shift 2]

qb=qf(1−
1
√k
)
qb=qf(1−
1
k
)
qb=qf(1+
1
√k
)
qb=qf(1+
1
k
)

Validate

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