Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q 1 coulomb and Q 2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is

q ( Q 2 − Q 1 ) ( 2 − 1 ) 2 . 4 π ε 0 R

q ( Q 1 + Q 2 ) ( 2 + 1 ) 2 . 4 π ε 0 R

Physics - Electrostatic Potential Capacitance - Quiz-6

A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it as the first combination has. The value of C₂ in terms of C₁ is:

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$\frac{2 C_{1}}{n_{1} n_{2}}$
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16 $\frac{n_{2}}{n_{1}}$ C₁
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2 $\frac{n_{2}}{n_{1}}$ C₁
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$\frac{16 C_{1}}{n_{1} {n_{2}}_{}}$

Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities $σ, - σ$ and $σ$ respectively. If $V_{A}, V_{B}$ and $V_{C}$ denote the potential of the three shells, if c=a+b, we have

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$V_{C} = V_{A} \neq V_{B}$
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$V_{C} = V_{B} \neq V_{A}$
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$V_{C} \neq V_{B} \neq V_{A}$
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$V_{C} = V_{B} = V_{A}$

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

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$\frac{C}{3}, \frac{V}{3}$
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$3 C, \frac{V}{3}$
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$\frac{C}{3}, 3 V$
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$3 C, 3 V$

The electric potential at a point (x,y,z) is given by

$V = - x^{2} y - x z^{3} + 4$

The electric field $\vec{E}$ at that point is

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() $\vec{E} = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}$
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() $\vec{E} = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})$
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() $\vec{E} = \hat{i} z^{3} + \hat{j} x y z + \hat{k} z^{2}$
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() $\vec{E} = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x$

The mean free path of electrons in a metal is $4 \times 10^{- 8} m .$ The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of $V m^{- 1}$ :

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$8 \times 10^{7}$
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$5 \times 10^{- 11}$
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$8 \times 10^{- 11}$
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$5 \times 10^{7}$

Two dielectric slabs of constant K₁ and K₂ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor

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$\frac{2 ε_{0} A}{2} (K_{1} + K_{2})$
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$\frac{2 ε_{0} A}{2} (\frac{K_{1} + K_{2}}{K_{1} \times K_{2}})$
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$\frac{2\epsilon _{0}A}{d}(\frac{k_{1}+k_{2}}{k_{1}-k_{2}})$
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$\frac{2 ε_{0} A}{d} (\frac{K_{1} \times K_{2}}{K_{1} + K_{2}})$

What is the equivalent capacitance between A and B in the given figure (all are in farad)

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$\frac{13}{18} F$
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$\frac{48}{13} F$
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$\frac{1}{31} F$
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$\frac{240}{71} F$

100 capacitors each having a capacity of 10 μF are connected in parallel and are charged by a potential difference of 100 kV. The energy stored in the capacitors and the cost of charging them, if electrical energy costs 108 paise per kWh, will be?

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10⁷ joule and 300 paise
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5 × 10⁶ joule and 300 paise
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5 × 10⁶ joule and 150 paise
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10⁷ joule and 150 paise

Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance between points x and y is (in μF)

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$\frac{5}{6}$
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$\frac{7}{6}$
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$\frac{8}{3}$
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2

A 10 μF capacitor and a 20 μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor

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$\frac{400}{9} V$
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$\frac{800}{9} V$
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400 V
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200 V

Two condensers C₁ and C₂ in a circuit are joined as shown in figure. The potential of point A is V₁ and that of B is V₂. The potential of point D will be

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

The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that

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Their capacities are equal
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Their capacities are 1 μF and 2 μF
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Their capacities are 0.5 μF and 1 μF
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Their capacities are infinite

In the given network capacitance, C₁ = 10 μF, C₂ = 5 μF and C₃ = 4 μF. What is the resultant capacitance between A and B

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2.2 μF
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3.2 μF
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1.2 μF
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4.7 μF

The equivalent capacitance between A and B is:

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2 μF
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3 μF
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5 μF
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0.5 μF

In the circuit shown in figure, each capacitor has a capacity of 3 μF. The equivalent capacity between A and B is

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$\frac{3}{4} μ F$
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3 μF
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6 μF
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5 μF

In the figure, three capacitors each of capacitance 6 pF are connected in series. The total capacitance of the combination will be

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9 × 10^–12 F
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6 × 10^–12 F
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3 × 10^–12 F
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2 × 10^–12 F

Equivalent capacitance between A and B is

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8 μF
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6 μF
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26 μF
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10/3 μF

In the figure a capacitor is filled with dielectrics. The resultant capacitance is

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$\frac{2 ε_{0} A}{d} [\frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}}]$
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$\frac{ε_{0} A}{d} [\frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}}]$
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$\frac{2 ε_{0} A}{d} [k_{1} + k_{2} + k_{3}]$
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None of these

Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100V. The charge on 15 μF is

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50 μC
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100 μC
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200 μC
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(4.) 280 μC

Two capacitors C₁ = 2 μF and C₂ = 6 μF in series, are connected in parallel to a third capacitor C₃ = 4 μF. This arrangement is then connected to a battery of e.m.f. = 2V, as shown in the figure. How much energy is lost by the battery in charging the capacitors.

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22 × 10^–6 J
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11 × 10^–6 J
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$(\frac{32}{3}) \times 10^{- 6} J$
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$(\frac{16}{3}) \times 10^{- 6} J$

A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constants K₁ and K₂ its capacity becomes C₁. The ratio of C₁ to C is

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K₁ + K₂
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$\frac{K_{1} K_{2}}{K_{1}-K_{2}}$
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$\frac{K_{1}+K_{2}}{K_{1} K_{2}}$
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$\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}$

The equivalent capacitance in the circuit between A and B will be

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1 μF
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2 μF
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3 μF
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$\frac{1}{3} μ F$

The equivalent capacitance between A and B is

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$\frac{C}{4}$
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$\frac{3 C}{4}$
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$\frac{C}{3}$
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$\frac{4 C}{3}$

In the given figure the capacitors C₁, C₃, C₄, C₅ have a capacitance 4 μF each and if the capacitor C₂ has a capacitance 10 μF, then effective capacitance between A and B will be

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2 μF
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4 μF
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6 μF
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8 μF

Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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$\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})$
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$\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$
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$\frac{1}{4} C {(V_{1} - V_{2})}^{2}$
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$\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Three capacitors each of capacity 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by

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Connecting them in parallel
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Connecting two in series and one in parallel
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Connecting two in parallel and one in series
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Connecting all of them in series

Three capacitors of capacitance 3 μF are connected in a circuit. Then their maximum and minimum capacitances will be

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9 μF, 1 μF
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8 μF, 2 μF
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9 μF, 0 μF
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3 μF, 2 μF

A capacitor of capacity C₁ is charged upto V volt and then connected to an uncharged capacitor of capacity C₂. Then final potential difference across each will be

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$\frac{C_{2} V}{C_{1} + C_{2}}$
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$(1 + \frac{C_{2}}{C_{1}}) V$
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$\frac{C_{1} V}{C_{1} + C_{2}}$
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$(1 - \frac{C_{2}}{C_{1}}) V$

Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 μC. The value of C₁ is

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2.5 μF
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15 μF
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1.5 μF
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0.1 μF

Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q₁ coulomb and Q₂ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is

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Zero
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$\frac{q (Q_{2} - Q_{1}) (\sqrt{2} - 1)}{\sqrt{2} . 4 π ε_{0} R}$
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$\frac{q \sqrt{2} (Q_{1} + Q_{2})}{4 π ε_{0} R}$
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$\frac{q (Q_{1} + Q_{2}) (\sqrt{2} + 1)}{\sqrt{2} . 4 π ε_{0} R}$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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