COMEDK 2023 Morning Shift | Capacitor Question 5 | Physics

A capacitor of capacity $$2 ~\mu \mathrm{F}$$ is charged upto a potential $$14 \mathrm{~V}$$ and then connected in parallel to an uncharged capacitor of capacity $$5 ~\mu \mathrm{F}$$. The final potential difference across each capacitor will be

6 V

4 V

8 V

14 V

COMEDK 2023 Morning Shift

MCQ (Single Correct Answer)

-0

A dielectric of dielectric constant $$K$$ is introduced such that half of its area of a capacitor of capacitance $$C$$ is occupied by it. The new capacity is

$$2 C$$

$$\frac{C}{2}$$

$$\frac{(1+K) C}{2}$$

$$2 C(1+K)$$

COMEDK 2023 Morning Shift

MCQ (Single Correct Answer)

-0

In the figure below, the capacitance of each capacitor is $$3 \mu \mathrm{F}$$. The effective capacitance between $$A$$ and $$B$$ is

COMEDK 2023 Morning Shift Physics - Capacitor Question 6 English

$$\frac{3}{4} \mu \mathrm{F}$$

$$3 \mu \mathrm{F}$$

$$6 \mu \mathrm{F}$$

$$5 \mu \mathrm{F}$$

COMEDK 2023 Morning Shift

MCQ (Single Correct Answer)

-0

If $$C$$ be the capacitance and $$V$$ be the electric potential, then the dimensional formula of $$\mathrm{CV}^2$$ is

$$\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^0\right]$$

$$\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-1}\right]$$

$$\left[\mathrm{M}^0 \mathrm{LT}^{-2} \mathrm{~A}^0\right]$$

$$\left[\mathrm{ML}^{-3} \mathrm{TA}\right]$$

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