For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is \frac{3}{4} d, where d is the separation between the plates of parallel plate capacitor. The new capacitance (C^{'}) in terms of original capacitance (C_{0}) is given by the following relation [16 Mar 2021 Shift 1]

Correct Answer is : C′=4KK+3C0

JEE Main 16 March 2021 Shift 1 Solved Paper

Section: Physics

Question : 2 of 90

Marks: +1, -0

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is ‌

d, where d is the separation between the plates of parallel plate capacitor. The new capacitance (C′) in terms of original capacitance (C0) is given by the following relation

[16 Mar 2021 Shift 1]

C′=‌
3+K
4K
C0
C′=‌
4+K
3
C0
C′=‌
4K
K+3
C0
C′=‌
4
3+K
C0

Validate

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