The insulated spheres of radii R 1 and R 2 having charges Q 1 and Q 2 respectively are connected to each other. There is -

A decrease in the energy of the system unless Q 1 R 2 = Q 2 R 1

No change in the energy of the system

An increase in the energy of the system

Always a decrease in the energy of the system

Physics - Electrostatic Potential Capacitance - Quiz-3

Equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are

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Planes parallel to yz-plane
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Planes parallel to xy-plane
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Planes parallel to xz-plane
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Coaxial cylinders of increasing radii around the x-axis

A bullet of mass 2 gm is having a charge of 2 μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s ?

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5 kV
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50 kV
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5 V
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50 V

In a certain charge distribution, all points having zero potential can be joined by a circle S. Points inside S have positive potential, and points outside S have a negative potential. A positive charge, which is free to move, is placed inside S .

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It will remain in equilibrium
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It can move inside S, but it cannot cross S
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It must cross S at some time
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It may move, but will ultimately return to its starting point

A square of side ‘a’ has charge Q at its centre and charge ‘q’ at one of the corners. The work required to be done in moving the charge ‘q’ from the corner to the diagonally opposite corner is -

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Zero
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$\frac{Q q}{4 π \in_{0} a}$
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$\frac{Q q \sqrt{2}}{4 π \in_{0} a}$
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$\frac{Q q}{2 π \in_{0} a}$

As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge –Q from the point A [co-ordinates (0, a)] to another point B [co-ordinates (a, 0)] along the straight path AB is

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Zero
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$(\frac{- q Q}{4 π ε_{0}} \frac{1}{a^{2}}) \sqrt{2} a$
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$(\frac{q Q}{4 π ε_{0}} \frac{1}{a^{2}}) \frac{a}{\sqrt{2}}$
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$(\frac{q Q}{4 π ε_{0}} \frac{1}{a^{2}}) \sqrt{2} a$

Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q₃is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac{q_{3}}{4 π ε_{0}} k$ , where k is

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8 q₂
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8 q₁
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6 q₂
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6 q₁

A hollow metallic sphere of radius R is given a charge Q. Then the potential at the centre is -

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Zero
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$\frac{1}{4 π ε_{0}} . \frac{Q}{R}$
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$\frac{1}{4 π ε_{0}} . \frac{2 Q}{R}$
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$\frac{1}{4 π ε_{0}} . \frac{Q}{2 R}$

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in -

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Reduction of charge on the plates and increase of potential difference across the plates
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Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates
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Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
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None of the above

The capacity of a condenser is 4 × 10^–6 farad and its potential is 100 volts. The energy released on discharging it fully will be -

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0.025 Joule
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0.05 Joule
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0.02 Joule
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0.04 Joule

The insulated spheres of radii R₁ and R₂ having charges Q₁and Q₂ respectively are connected to each other. There is -

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No change in the energy of the system
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An increase in the energy of the system
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Always a decrease in the energy of the system
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A decrease in the energy of the system unless Q₁R₂ = Q₂R₁

The energy stored in a condenser of capacity C which has been raised to a potential V is given by -

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

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

If two conducting spheres are separately charged and then brought in contact -

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The total energy of the two spheres is conserved

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The total charge on the two spheres is conserved

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Both the total energy and charge are conserved

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The final potential is always the mean of the original potentials of the two spheres

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

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8 times

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4 times

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2 times

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32 times

A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant ‘k’ of the oil is

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0.45

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0.55

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1.10

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2.20

Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes -

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$\frac{ε_{0} A}{d + t (1 - \frac{1}{k})}$

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$\frac{ε_{0} A}{d + t (1 + \frac{1}{k})}$

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$\frac{ε_{0} A}{d - t (1 - \frac{1}{k})}$

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$\frac{ε_{0} A}{d - t (1 + \frac{1}{k})}$

The capacity of parallel plate condenser depends on

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The type of metal used

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The thickness of plates

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The potential applied across the plates

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The separation between the plates

The capacity of a parallel plate condenser is C. It's capacity when the separation between the plates is halved will be?

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4 C

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2 C

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$\frac{C}{2}$

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$\frac{C}{4}$

Eight small drops, each of radius r and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

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8 : 1

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4 : 1

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2 : 1

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1 : 8

1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of

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1000

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100

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10

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1

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

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Increased proportional to 2

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

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Increased proportional to $\sqrt{2}$

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

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:

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high k and high x.

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high k and low x.

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low k and low x.

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low k and high x.

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

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Decreases K times

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Increases K times

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Increases K² times

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Remains constant

The capacity of a parallel plate capacitor increases with the

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Decrease of its area

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Increase of its distance

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Increase of its area

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None of the above

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

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From P to Q

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From Q to P

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

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The information is incomplete

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

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

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

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

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

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

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The electric intensity between the plates will decrease

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The electric intensity between the plates will increase

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The electric intensity between the plates will remain unchanged

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The P.D. between the plates will decrease

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 the glass (dielectric constant = 5) between the plates, the capacity of the condenser and the energy stored in it will respectively be -

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

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

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

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

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

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$\frac{q^{2}}{2 ε_{0} A K}$

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$\frac{q^{2}}{ε_{0} A K}$

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$\frac{q}{2 ε_{0} A}$

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$\frac{q^{2}}{2 ε_{0} A^{2} K}$

A 6 μF capacitor is charged from 10 volts to 20 volts. Increase in energy will be

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18 × 10^–4 J

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9 × 10^–4 J

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4.5 × 10^–4 J

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9 × 10^–6 J

Twenty seven drops of water of the same size are equally and similarly charged. They are then united to form a bigger drop. By what factor will the electrical potential changes

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9 times

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27 times

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6 times

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3 times

0:0:1

0:0:1

Practice Physics Quiz Questions and Answers

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