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

The charge on the capacitor increases

The voltage across the plates decreases

The capacitance increases

The electrostatic energy stored in the capacitor increases

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Explanation

When the battery is disconnected, the charge will remain the same in any case.

The capacitance of a parallel plate capacitor is given by $C = \frac{ε_{0} A}{d}$

When d is increased, the capacitance will decrease and because the charge remains the same, so according to q = CV, the voltage will increase, Hence the electrostatic energy stored in the capacitor will increase.

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A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed respectively will be

Zero

$4 π ε_{0} a$

$4 π ε_{0} b$

$4 π ε_{0} a (\frac{b}{b - a})$

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Explanation

Capacity when outer sphere is earthed

$C_{1} = 4 π ε_{0} \frac{a b}{b - a}$

Capacity when inner sphere is earthed

$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} a b}{b - a} = 4 π ε_{0} (\frac{b^{2}}{b - a})$

Difference in capacity = C₂ – C₁ = $4 π ε_{0} b$

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The intensity of electric field at a point between the plates of a charged capacitor

Is directly proportional to the distance between the plates

Is inversely proportional to the distance between the plates

Is inversely proportional to the square of the distance between the plates

Does not depend upon the distance between the plates

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Explanation

Electric field between the plates of parallel plate capacitor is uniform and it doesn't depend upon distance.

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In a spherical condenser radius of the outer sphere is R. The difference in the radii of the outer and inner sphere is x. Its capacity is proportional to

$\frac{x R}{(R - x)}$

$\frac{x (R - x)}{r}$

$\frac{R (R - x)}{x}$

$\frac{R}{x}$

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Explanation

$C \propto \frac{a b}{b - a};$ a = R – x, b = R so, $C \propto \frac{R (R - x)}{x}$

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A capacitor, when filled with a dielectric K = 3, has charge Q₀, voltage V₀ and field E₀. If the dielectric is replaced with another one having K = 9 the new values of charge, voltage and field will be respectively, when the capacitor is not connected with a battery -

$3 Q_{0}, 3 V_{0}, 3 E_{0}$

$Q_{0}, 3 V_{0}, 3 E_{0}$

$Q_{0}, \frac{V_{0}}{3}, 3 E_{0}$

$Q_{0}, \frac{V_{0}}{3}, \frac{E_{0}}{3}$

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Explanation

When there is no battery, the charge remains the same while potential difference and electric field decreases as the capacitance increases.

i.e.

$C' = \frac{9}{3} \times \frac{ε_{0} A}{d} = 3 C_{0}$

$Q' = Q_{0} = C_{0} V_{0}, V' = \frac{Q'}{C_{0}} = \frac{C_{0} V_{0}}{3 C_{0}} = \frac{V_{0}}{3}$

and $E' = \frac{V'}{d} = \frac{V_{0}}{3 d} = \frac{E_{0}}{3}$

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A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes

$\sqrt{2} C$

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

$\frac{C}{2}$

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Explanation

$C_{a i r} = \frac{C_{m e d i u m}}{K} = \frac{C}{2}$

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What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ?

$1 .694 \times 10^{9} m^{2}$

$4 .529 \times 10^{9} m^{2}$

$9 .281 \times 10^{9} m^{2}$

$12 .981 \times 10^{9} m^{2}$

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Explanation

We have $C = \frac{ε_{0} A}{d}$

⇒ $A = \frac{C d}{ε_{0}} = \frac{3 \times 5 \times 10^{- 3}}{8 .85 \times 10^{- 12}}$

$= 1 .7 \times 10^{9} m^{2}$

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The energy stored in the condenser is

$\frac{1}{2} Q V$

$\frac{1}{2} C$

$\frac{1}{2} \frac{Q}{C}$

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Explanation

The energy stored $= \frac{1}{2} Q V$

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Two identical charged spherical drops each of capacitance C merge to form a single drop. The resultant capacitance is -

Equal to 2C

Greater than 2C

Less than 2C but greater than C

Less than C

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Explanation

$C' = n^{1 / 3} C$

⇒ $C' = 2^{1 / 3} C$

⇒ 2C > C' > C

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A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor ?

1.5 × 10^–8Joule

2.5 × 10^–7Joule

3.5 × 10^–5Joule

4.5 × 10^–2Joule

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 12 \times 10^{- 12} \times {(50)}^{2}$ = 1.5 × 10^–8 J

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