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A parallel plate condenser is immersed in an oil of dielectric constant 2. The field between the plates is

Increased proportional to 2

Decreased proportional to $\frac{1}{2}$

Increased proportional to $\sqrt{2}$

Decreased proportional to $\frac{1}{\sqrt{2}}$

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Explanation

$E_{medium} = \frac{E_{air}}{K} = \frac{E}{2}$

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If the dielectric constant and dielectric strength be denoted by k and x respectively, then a material suitable for use as a dielectric in a capacitor must have

High k and high x

High k and low x

Low k and low x

Low k and high x

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Explanation

High K means good insulating property and high x means able to withstand electric field to a higher value.

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When air in a capacitor is replaced by a medium of dielectric constant K, the capacity -

Decreases K times

Increases K times

Increases K² times

Remains constant

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Explanation

$C_{m e d i u m} = K \times C_{a i r}$

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64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small drop is q, then the charge on the big drop will be

16q

64q

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Explanation

By using Q = nq ⇒ Q = 64q

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The capacity of a parallel plate capacitor increases with the

Decrease of its area

Increase of its distance

Increase of its area

None of the above

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Explanation

Capacity of parallel plate capacitor $C = \frac{ε_{0} A}{d}$

⇒ C ∝ A

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The radii of two metallic spheres P and Q are r₁ and r₂ respectively. They are given the same charge. If r₁ > r₂ , then on connecting them with a thin wire, the charge will flow

From P to Q

From Q to P

Neither the charge will flow from P to Q nor from Q to P

The information is incomplete

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Explanation

Since charge flows from high potential to lower potential.

If positive charge is given, then V₁ < V₂ as r₁ > r₂

So positive charge flows from Q → P

If negative charge is given, then V₁ > V₂

So negative charge flows from P → Q.

Since it is not given that whether the charge given is positive or negative, hence the information is incomplete.

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Between the plates of a parallel plate condenser, a plate of thickness t₁ and dielectric constant k₁ is placed. In the rest of the space, there is another plate of thickness t₂ and dielectric constant k₂. The potential difference across the condenser will be

$\frac{Q}{A ε_{0}} (\frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}})$

$\frac{ε_{0} Q}{A} (\frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}})$

$\frac{Q}{A ε_{0}} (\frac{k_{1}}{t_{1}} + \frac{k_{2}}{t_{2}})$

$\frac{ε_{0} Q}{A} (k_{1} t_{1} + k_{2} t_{2})$

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Explanation

Potential difference across the condenser

$V = V_{1} + V_{2} = E_{1} t_{1} + E_{2} t_{2} = \frac{σ}{K_{1} ε_{0}} t_{1} + \frac{σ}{K_{2} ε_{0}} t_{2}$

⇒ $V = \frac{σ}{ε_{0}} (\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}}) = \frac{Q}{A ε_{0}} (\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}})$

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The true statement is, on increasing the distance between the plates of a parallel plate condenser -

The electric intensity between the plates will decrease

The electric intensity between the plates will increase

The electric intensity between the plates will remain unchanged

The P.D. between the plates will decrease

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Explanation

Electric field between the plates of a parallel plate capacitor $E = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}}$ i.e. $E \propto d^{o}$

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The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C_O and W_O. If the air is replaced by glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be -

$5 C_{o}, 5 W_{o}$

$5 C_{o}, \frac{W_{0}}{5}$

$\frac{C_{o}}{5}, 5 W_{o}$

$\frac{C_{o}}{5}, \frac{W_{o}}{5}$

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Explanation

When a dielectric K is introduced in a parallel plate condenser its capacity becomes K times. Hence C’ = 5C₀. Energy stored $W_{0} = \frac{q^{2}}{2 C_{0}}$

∴ $W' = \frac{q^{2}}{2 C'} = \frac{q^{2}}{2 \times 5 C_{0}}$ ⇒ $W' = \frac{W_{0}}{5}$

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Force of attraction between the plates of a parallel plate capacitor whose dielectric constant is K will be -

$\frac{q^{2}}{2 ε_{0} A K}$

$\frac{q^{2}}{ε_{0} A K}$

$\frac{q}{2 ε_{0} A}$

$\frac{q^{2}}{2 ε_{0} A^{2} K}$

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Explanation

Force on one plate due to another is

F = qE = $q \times \frac{σ}{2 ε_{0} K} = q (\frac{q}{2 A K ε_{0}}) = \frac{q^{2}}{2 A K ε_{0}}$

(where $\frac{σ}{2 ε_{0} K}$ is the electric field produced by one plate at the location of other).

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