Physics - Electrostatic Potential Capacitance - Quiz-1

If W be the amount of heat produced in the process of charging an uncharged capacitor then the amount of energy stored in it is

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2W
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$\frac{W}{2}$
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W
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zero

A metallic sphere of capacitance $C_{1}$ , charged to electric potential $V_{1}$ is connected by a metal wire to another metallic sphere of capacitance $C_{2}$ charged to electric potential $V_{2}$ . The amount of heat produced in the connecting wire during the process is

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$\frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} + V_{2})}^{2}$
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$\frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$
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$\frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$
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zero

The electric potential at the surface of a charged solid sphere of insulator is 20V. The value of electric potential at its centre will be

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30V
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20V
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40V
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Zero

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

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$\frac{K C}{2 (K + 1)}$
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$\frac{2 K C}{K + 1}$
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$\frac{5 K C}{4 K + 1}$
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$\frac{4 K C}{3 K + 1}$

The electric potential at a point at distance 'r' from a short dipole is proportional to

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$r^{2}$
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$r^{- 1}$
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$r^{- 2}$
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$r^{1}$

A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is

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$\frac{3 V}{19 R}$
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$\frac{V}{6 R}$
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$\frac{3 V}{32 R}$
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$\frac{3 V}{16 R}$

Capacitors $C_{1} = 10 μ F a n d C_{2} = 30 μ F$ are connected in series across a source of emf 20KV. The potential difference across $C_{1}$ will be

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5 KV
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15 KV
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10 KV
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20 KV

A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown. Let $V_{A}, V_{B}, V_{C}$ be the potentials at points A, B and C respectively. Then

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$V_{A} < V_{B} < V_{C}$
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$V_{A} > V_{B} > V_{C}$
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$V_{C} > V_{B} = V_{A}$
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$V_{A} = V_{B} = V_{C}$

Four particles each having charge q are placed at the vertices of a square of side a. The value of the electric potential at the midpoint of one of the side will be

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0
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$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 + \frac{2}{\sqrt{5}})$
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$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 - \frac{2}{\sqrt{5}})$
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$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (1 + \frac{1}{\sqrt{5}})$

If E be the electric field inside a parallel plate capacitor due to Q and -Q charges on the two plates, then electrostatic force on plate having charge -Q due to the plate having charge +Q will be

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-QE
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$\frac{- Q E}{2}$
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QE
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$\frac{- Q E}{4}$

Two metallic spheres of radii 2cm and 3cm are given charges 6mC and 4mC respectively. The final charge on the smaller sphere will be if they are connected by a conducting wire

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4mC
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6mC
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5mC
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10mC

When a proton at rest is accelerated by a potential difference V, its speed is found to be v. The speed of an $α$ -particle when accelerated by the same potential difference from rest will be:

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v
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$\frac{v}{\sqrt{2}}$
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$v \sqrt{2}$
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2v

In the circuit shown in figure, energy stored in 6 $μF$ capacitor will be

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$48 \times 10^{- 6} J$
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$32 \times 10^{- 6} J$
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$96 \times 10^{- 6} J$
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$24 \times 10^{- 6} J$

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by

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200 V/m, making an angle $120^{0}$ with the x-axis
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100 V/m, pointing towards the negative x-axis
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200 V/m, making an angle $- 60^{0}$ with the x-axis
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100 V/m, making an angle $30^{0}$ with the x-axis

An air capacitor of capacity $C = 10 μ F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is

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120 $μ C$
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699 $μ C$
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480 $μ C$
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24 $μ C$

A and B are two concentric metallic shells. If A is positively charged and B is earthed, then electric

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Field at common centre is non-zero
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Field outside B is nonzero
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Potential outside B is positive
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Potential at common centre is positive

An elementary particle of mass m and charge e is projected with velocity v at a much more massive particle of charge Ze, where $Z > 0$ . What is the closest possible approach of the incident particle ?

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$\frac{Z e^{2}}{2 π ε_{0} m v^{2}}$
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$\frac{Z e}{4 π ε_{0} m v^{2}}$
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$\frac{Z e^{2}}{8 π ε_{0} m v^{2}}$
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$\frac{Z e}{8 π ε_{0} m v^{2}}$

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is

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0
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$\frac{\sqrt{2} Q^{2}}{4 π ε_{0} a}$
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$\frac{\sqrt{2} Q^{2}}{π ε_{0} a}$
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$\frac{Q^{2}}{2 π ε_{0} a}$

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is

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a/b
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b/a
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a²/b²
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b²/a²

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be

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$V \sqrt{e / m}$
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$\sqrt{e V / m}$
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$\sqrt{2 e V / m}$
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$2 e V / m$

The dimension of (1/2) $ε_{0} E^{2} (ε_{0}$ : permittivity of free space; E: electric field) is

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MLT^–1
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ML²L^–2
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ML^–1T^–2
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ML²T^–1

A table tennis ball that has been covered with conducting paint is suspended by a silk thread so that it hangs between two plates, out of which one is earthed and other is connected to a high voltage generator. This ball

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Is attracted towards high voltage plate and stays there
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Hangs without moving
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Swing backward and forward hitting each plate in turn
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Is attracted to the earthed plate and stays there

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by (r >> 2a) (Where p = 2qa)

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$V = \frac{p \cos θ}{4 π ε_{0} r^{2}}$
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$V = \frac{p \cos θ}{4 π ε_{0} r}$
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$V = \frac{p \sin θ}{4 π ε_{0} r}$
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$V = \frac{p \cos θ}{2 π ε_{0} r^{2}}$

A charge +q is fixed at each of the points $x = x_{0}, x = 3 x_{0}, x = 5 x_{0}$ ..... infinite, on the x-axis, and a charge –q is fixed at each of the points $x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0}$ ,..... infinite. Here x₀ is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be $\frac{Q}{4 π ε_{0} r}$ . Then, the potential at the origin due to the above system of charges is:

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0
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$\frac{q}{8 π ε_{0} x_{0} \ln 2}$
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∞
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$\frac{q \log 2}{4 π ε_{0} x_{0}}$

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing co-parallel to the positive direction of the x-axis. The coordinates of the points P, Q, R, and S are $(a, b, 0), (2 a, 0, 0), (a, - b, 0)$ and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression

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qEa
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– qEa
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$q E a \sqrt{2}$
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$q E \sqrt{[{(2 a)}^{2} + b^{2}]}$

Inside a hollow charged spherical conductor, the potential -

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Is constant
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Varies directly as the distance from the centre
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Varies inversely as the distance from the centre
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Varies inversely as the square of the distance from the centre

Two small spheres each carrying a charge q are placed r meter apart. If one of the spheres is taken around the other one in a circular path of radius r, the work done will be equal to

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Force between them × r
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Force between them × 2πr
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Force between them / 2πr
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Zero

Two charged spheres of radii 10 cm and 15 cm are connected by a thin wire. No current will flow, if they have -

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The same charge on each
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The same potential
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The same energy
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The same field on their surfaces

A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p which is at a distance $\frac{R}{2}$ from the centre of the shell is

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$\frac{(q + Q)}{4 π ε_{0}} \frac{2}{R}$
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$\frac{2 Q}{4 π ε_{0} R}$
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$\frac{2 Q}{4 π ε_{0} R} - \frac{2 q}{4 π ε_{0} R}$
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$\frac{2 Q}{4 π ε_{0} R} + \frac{q}{4 π ε_{0} R}$

The electric potential V at any point O (x, y, z all in metres) in space is given by $V = 4 x^{2} v o l t$ . The electric field at the point $(1 m, 0, 2 m)$ in volt/metre is -

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8 along negative x-axis
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8 along positive x-axis
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16 along negative x-axis
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16 along positive z-axis

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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