MHT CET 2023 9th May Morning Shift | Capacitor Question 44 | Physics

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $$C_1$$ and $$C_2$$ with their capacitance in the ratio $$1: 2$$ so that energy stored in these two cases becomes same is

$$3: \sqrt{2}$$

$$\sqrt{2}: 3$$

$$2: 9$$

$$9: 2$$

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

-0

The potential energy of charged parallel plate capacitor is $$v_0$$. If a slab of dielectric constant $$\mathrm{K}$$ is inserted between the plates, then the new potential energy will be

$$\frac{v_0}{\mathrm{~K}}$$

$$v_0 \mathrm{~K}^2$$

$$\frac{v_0}{\mathrm{~K}^2}$$

$$\mathrm{v}_0^2$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

-0

A parallel plate air capacitor has a uniform electric field 'E' in the space between the plates. Area of each plate is A and the distance between the plates is '$$\mathrm{d}$$'. The energy stored in the capacitor is $$\left[\varepsilon_0=\right.$$ permittivity of free space)

$$2 \varepsilon_0 \mathrm{EAd}$$

$$\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$$

$$\frac{\varepsilon_0 \mathrm{E}^2}{2 \mathrm{Ad}}$$

$$\frac{\mathrm{E}^2 \mathrm{Ad}}{2 \varepsilon_0}$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

-0

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then

the capacitance of the capacitor increases

the charge on the capacitor decreases

the voltage across the capacitor increases

the energy stored in the capacitor decreases

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