Physics - Electrostatic Potential Capacitance - Quiz-4

The dielectric constant k of an insulator cannot be

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3
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6
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8
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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

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The charge on the capacitor increases
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The voltage across the plates decreases
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The capacitance increases
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The electrostatic energy stored in the capacitor increases

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

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Zero
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$4 π ε_{0} a$
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$4 π ε_{0} b$
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$4 π ε_{0} a (\frac{b}{b - a})$

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

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Is directly proportional to the distance between the plates
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Is inversely proportional to the distance between the plates
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Is inversely proportional to the square of the distance between the plates
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Does not depend upon the distance between the plates

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

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

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 -

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

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

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

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

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Equal to 2C
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Greater than 2C
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Less than 2C but greater than C
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Less than C

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

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(1) 5 × 10^–8Joule
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(2) 5 × 10^–7Joule
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(3) 5 × 10^–5Joule
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(4) 5 × 10^–2Joule

The capacity of a parallel plate condenser is 15 μF, when the distance between its plates is 6 cm. If the distance between the plates is reduced to 2 cm, then the capacity of this parallel plate condenser will be?

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15 μF
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30 μF
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45 μF
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60 μF

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is

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(2) 15.6 μF
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(1) 12 μF
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(3) 19.2 μF
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(4) 22.4 μF

Two insulated metallic spheres of 3 μF and 5 μF capacitances are charged to 300 V and 500V respectively. The energy loss, when they are connected by a wire is

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0.012 J
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0.0218 J
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0.0375 J
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3.75 J

A charge of 40 μC is given to a capacitor having capacitance C = 10 μF. The stored energy in ergs is

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80 × 10^–6
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800
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80
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8000

A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V₀. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

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$\frac{3 ε_{0} A V_{0}^{2}}{d}$
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$\frac{ε_{0} A V_{0}^{2}}{2 d}$
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$\frac{ε_{0} A V_{0}^{2}}{3 d}$
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$\frac{ε_{0} A V_{0}^{2}}{d}$

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is E₀. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant K without disturbing the battery connections, the field between the plates shall be

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K E₀
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E₀
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$\frac{E_{0}}{K}$
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None of the above

If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance

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Decreases two times
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Increases two times
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Increases four times
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Remain the same

If there are n capacitors each of capacitance C in parallel connected to V volt source, then the energy stored is equal to

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CV
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$\frac{1}{2} n C V^{2}$
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CV²
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$\frac{1}{2 n} C V^{2}$

The unit of electric permittivity is

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Volt/m²
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Joule/coulomb
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Farad/m
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Henry/m

The work done in placing a charge of 8 × 10^–18 coulomb on a condenser of capacity 100 micro-farad is

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32× 10^–32Joule
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16 × 10^–32Joule
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3.1 × 10^–26Joule
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4 × 10^–10Joule

A parallel plate capacitor of capacity C₀ is charged to a potential V₀

(i) The energy stored in the capacitor when the battery is disconnected and the separation is doubled is E₁

(ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂.

Then E₁ / E₂ value is :

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4
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3/2
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2
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1/2

As in figure shown, if a capacitor C is charged by connecting it with resistance R, then energy given by the battery will be

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$\frac{1}{2} C V^{2}$
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More than $\frac{1}{2} C V^{2}$
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Less than $\frac{1}{2} C V^{2}$
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Zero

Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other

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The charges on the free plates are destroyed
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The charges on the free plates are enhanced
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The energy stored in the system increases
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The potential difference in the free plates becomes 2V

A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

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C
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nC
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(n – 1)C
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(n + 1)C

Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is

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$\frac{2 ε_{0} A}{d}$
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$\frac{3 ε_{0} A}{d}$
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$\frac{4 ε_{0} A}{d}$
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$\frac{ε_{0} A}{d}$

A parallel plate capacitor with air as medium between the plates has a capacitance of 10 μF. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant k₁ = 2 and k₂ = 4. The capacitance of the system will now be

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10 μF
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20 μF
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30 μF
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40 μF

Three capacitors are connected to D.C. source of 100 volts shown in the adjoining figure. If the charge accumulated on plates of C₁, C₂ and C₃ are $q_{a}, q_{b}, q_{c}, q_{d} . q_{e}$ and q_f respectively, then

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$q_{b} + q_{d} + q_{f} = \frac{100}{9} C$
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$q_{b} + q_{d} + q_{f} = 0$
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$q_{a} + q_{c} + q_{e} = 50 C$
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$q_{b} = q_{d} = q_{f}$

n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

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Energy and potential difference remain the same
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Energy remains the same and the potential difference is nV
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Energy increases n times and potential difference is nV
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Energy increases n times and potential difference remains the same

Five capacitors of 10 μF capacity each are connected to a d.c. potential of 100 volts as shown in the adjoining figure. The equivalent capacitance between the points A and B will be equal to

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40 μF
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20 μF
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30 μF
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10 μF

Three capacitors of capacitances 3 μF, 9 μF and 18 μF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $(\frac{C_{s}}{C_{p}})$ will be:

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1 : 15
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15 : 1
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1 : 1
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1 : 3

Four condensers each of capacity 4 μF are connected as shown in figure and V_P – V_Q = 15 volts. The energy stored in the system is

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1800 ergs
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3600 ergs
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5400 ergs
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2400 ergs

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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