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PHXII01:ELECTRIC CHARGES AND FIELDS

358100 A point charge $Q$ is fixed at the centre of an insulated disc of mass $M$ . The disc rests on a rough horizontal plane. Another charge having same value as point charge is fixed vertically above the centre of the disc at a height of $h$ . After the disc is displaced slightly in the horizontal direction (friction is sufficient to prevent slipping). If the period of oscillation of disc is found to be $2 π \frac{h}{Q} \sqrt{n π ε_{0} M h}$ ,
supporting img

1 2

2 6

3 8

4 10

Explanation:

Let the radius of the disc be $R$ . If the disc is displaced $x$ , the corresponding angular displacement $θ = x / R$ .
The restoring torque $τ$ about the point of contact of disc with ground,
$τ_{p} = (F \sin θ) R$
$(F \sin θ)$ $R = I α = [\frac{M R^{2}}{2} + M R^{2}] α (1)$
where $F = \frac{Q^{2}}{4 π ε_{0} (h^{2} + x^{2})} (2)$

supporting img

and $\sin θ = \frac{x}{\sqrt{h^{2} + x^{2}}} (3)$
From (1), (2) and (3), we get
$\frac{Q^{2} x R}{4 π ε_{0} {(h^{2} + x^{2})}^{3 / 2}} = [\frac{M R^{2}}{2} + M R^{2}] α$
$o r α = \frac{Q^{2} x}{6 π ε_{0} M R {(h^{2} + x^{2})}^{3 / 2}}$
$F o r x << h, α ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
$a n d α = \frac{a}{R} ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other.
Thus, $a = - \frac{Q^{2} x}{6 π ε_{0} M h^{3}}$
Hence, $ω = \sqrt{\frac{Q^{2}}{6 π ε_{0} M h^{3}}}$
or $T = 2 π \sqrt{\frac{6 π ε_{0} M h^{3}}{Q^{2}}} = 2 π \frac{h}{Q} \sqrt{6 π ε_{0} M h}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358101 Two equal point charges each of $3 μ C$ are separated by a certain distance in metres. If they are located at $(\hat{i} + \hat{j} + \hat{k})$ and $(2 \hat{i} + 3 \hat{j} + 3 \hat{k}),$ the electrostatic force between them is

1

9 \times 10^{- 2} N

2

9 \times 10^{- 3} N

3

10^{- 3} N

4

9 \times 10^{3} N

Explanation:

Here, $q_{1} = 3 \times 10^{- 6} C, q_{2} = 3 \times 10^{- 6} C$
$r_{1} = \hat{i} + \hat{j} + \hat{k}, r_{2} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$
$∴ \vec{r} = \vec{r_{2}} - \vec{r_{1}}$
$= (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}$
The distance between the two charges is
$| \vec{r_{1}} | = \sqrt{{(1)}^{2} + {(2)}^{2} + {(2)}^{2}} = 3$
According to
coulomb’s
law,
$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6} \times 3 \times 10^{- 6}}{{(3)}^{2}} = 9 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358102 Two point charges exert on each other a force $F$ when they are placed $r$ distnce apart in air. If they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert double the force. The distance $R$ equals

1

\frac{r}{\sqrt{2 k}}

2

r k

3

\frac{r}{k}

4

r \sqrt{k}

Explanation:

In air medium the force is $F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
In a medium of dielectric constant $k$ , the force is
$F^{'} = \frac{q_{1} q_{2}}{4 π k ε R^{2}}$
As, $F^{'} = 2 F \Rightarrow \frac{q_{1} q_{2}}{4 π k ε_{0} R^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
$∴ R = \frac{r}{\sqrt{2 k}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358103 If point charges $Q_{1} = 2 \times 10^{- 7} C and Q_{2} = 3 \times 10^{- 7} C$ are at $30 cm$ separation, then find electrostatic force between them.

1

2 \times 10^{- 3} N

2

6 \times 10^{- 3} N

3

5 \times 10^{- 3} N

4

1 \times 10^{- 3} N

Explanation:

$∵$ Electrostatic force, $F = k \frac{Q_{1} Q_{2}}{r^{2}}$
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 7} \times 3 \times 10^{- 7}}{{(30 \times 10^{- 2})}^{2}}$
$= 6 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358100 A point charge $Q$ is fixed at the centre of an insulated disc of mass $M$ . The disc rests on a rough horizontal plane. Another charge having same value as point charge is fixed vertically above the centre of the disc at a height of $h$ . After the disc is displaced slightly in the horizontal direction (friction is sufficient to prevent slipping). If the period of oscillation of disc is found to be $2 π \frac{h}{Q} \sqrt{n π ε_{0} M h}$ ,
supporting img

1 2

2 6

3 8

4 10

Explanation:

Let the radius of the disc be $R$ . If the disc is displaced $x$ , the corresponding angular displacement $θ = x / R$ .
The restoring torque $τ$ about the point of contact of disc with ground,
$τ_{p} = (F \sin θ) R$
$(F \sin θ)$ $R = I α = [\frac{M R^{2}}{2} + M R^{2}] α (1)$
where $F = \frac{Q^{2}}{4 π ε_{0} (h^{2} + x^{2})} (2)$

supporting img

and $\sin θ = \frac{x}{\sqrt{h^{2} + x^{2}}} (3)$
From (1), (2) and (3), we get
$\frac{Q^{2} x R}{4 π ε_{0} {(h^{2} + x^{2})}^{3 / 2}} = [\frac{M R^{2}}{2} + M R^{2}] α$
$o r α = \frac{Q^{2} x}{6 π ε_{0} M R {(h^{2} + x^{2})}^{3 / 2}}$
$F o r x << h, α ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
$a n d α = \frac{a}{R} ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other.
Thus, $a = - \frac{Q^{2} x}{6 π ε_{0} M h^{3}}$
Hence, $ω = \sqrt{\frac{Q^{2}}{6 π ε_{0} M h^{3}}}$
or $T = 2 π \sqrt{\frac{6 π ε_{0} M h^{3}}{Q^{2}}} = 2 π \frac{h}{Q} \sqrt{6 π ε_{0} M h}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358101 Two equal point charges each of $3 μ C$ are separated by a certain distance in metres. If they are located at $(\hat{i} + \hat{j} + \hat{k})$ and $(2 \hat{i} + 3 \hat{j} + 3 \hat{k}),$ the electrostatic force between them is

1

9 \times 10^{- 2} N

2

9 \times 10^{- 3} N

3

10^{- 3} N

4

9 \times 10^{3} N

Explanation:

Here, $q_{1} = 3 \times 10^{- 6} C, q_{2} = 3 \times 10^{- 6} C$
$r_{1} = \hat{i} + \hat{j} + \hat{k}, r_{2} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$
$∴ \vec{r} = \vec{r_{2}} - \vec{r_{1}}$
$= (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}$
The distance between the two charges is
$| \vec{r_{1}} | = \sqrt{{(1)}^{2} + {(2)}^{2} + {(2)}^{2}} = 3$
According to
coulomb’s
law,
$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6} \times 3 \times 10^{- 6}}{{(3)}^{2}} = 9 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358102 Two point charges exert on each other a force $F$ when they are placed $r$ distnce apart in air. If they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert double the force. The distance $R$ equals

1

\frac{r}{\sqrt{2 k}}

2

r k

3

\frac{r}{k}

4

r \sqrt{k}

Explanation:

In air medium the force is $F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
In a medium of dielectric constant $k$ , the force is
$F^{'} = \frac{q_{1} q_{2}}{4 π k ε R^{2}}$
As, $F^{'} = 2 F \Rightarrow \frac{q_{1} q_{2}}{4 π k ε_{0} R^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
$∴ R = \frac{r}{\sqrt{2 k}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358103 If point charges $Q_{1} = 2 \times 10^{- 7} C and Q_{2} = 3 \times 10^{- 7} C$ are at $30 cm$ separation, then find electrostatic force between them.

1

2 \times 10^{- 3} N

2

6 \times 10^{- 3} N

3

5 \times 10^{- 3} N

4

1 \times 10^{- 3} N

Explanation:

$∵$ Electrostatic force, $F = k \frac{Q_{1} Q_{2}}{r^{2}}$
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 7} \times 3 \times 10^{- 7}}{{(30 \times 10^{- 2})}^{2}}$
$= 6 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358100 A point charge $Q$ is fixed at the centre of an insulated disc of mass $M$ . The disc rests on a rough horizontal plane. Another charge having same value as point charge is fixed vertically above the centre of the disc at a height of $h$ . After the disc is displaced slightly in the horizontal direction (friction is sufficient to prevent slipping). If the period of oscillation of disc is found to be $2 π \frac{h}{Q} \sqrt{n π ε_{0} M h}$ ,
supporting img

1 2

2 6

3 8

4 10

Explanation:

Let the radius of the disc be $R$ . If the disc is displaced $x$ , the corresponding angular displacement $θ = x / R$ .
The restoring torque $τ$ about the point of contact of disc with ground,
$τ_{p} = (F \sin θ) R$
$(F \sin θ)$ $R = I α = [\frac{M R^{2}}{2} + M R^{2}] α (1)$
where $F = \frac{Q^{2}}{4 π ε_{0} (h^{2} + x^{2})} (2)$

supporting img

and $\sin θ = \frac{x}{\sqrt{h^{2} + x^{2}}} (3)$
From (1), (2) and (3), we get
$\frac{Q^{2} x R}{4 π ε_{0} {(h^{2} + x^{2})}^{3 / 2}} = [\frac{M R^{2}}{2} + M R^{2}] α$
$o r α = \frac{Q^{2} x}{6 π ε_{0} M R {(h^{2} + x^{2})}^{3 / 2}}$
$F o r x << h, α ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
$a n d α = \frac{a}{R} ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other.
Thus, $a = - \frac{Q^{2} x}{6 π ε_{0} M h^{3}}$
Hence, $ω = \sqrt{\frac{Q^{2}}{6 π ε_{0} M h^{3}}}$
or $T = 2 π \sqrt{\frac{6 π ε_{0} M h^{3}}{Q^{2}}} = 2 π \frac{h}{Q} \sqrt{6 π ε_{0} M h}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358101 Two equal point charges each of $3 μ C$ are separated by a certain distance in metres. If they are located at $(\hat{i} + \hat{j} + \hat{k})$ and $(2 \hat{i} + 3 \hat{j} + 3 \hat{k}),$ the electrostatic force between them is

1

9 \times 10^{- 2} N

2

9 \times 10^{- 3} N

3

10^{- 3} N

4

9 \times 10^{3} N

Explanation:

Here, $q_{1} = 3 \times 10^{- 6} C, q_{2} = 3 \times 10^{- 6} C$
$r_{1} = \hat{i} + \hat{j} + \hat{k}, r_{2} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$
$∴ \vec{r} = \vec{r_{2}} - \vec{r_{1}}$
$= (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}$
The distance between the two charges is
$| \vec{r_{1}} | = \sqrt{{(1)}^{2} + {(2)}^{2} + {(2)}^{2}} = 3$
According to
coulomb’s
law,
$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6} \times 3 \times 10^{- 6}}{{(3)}^{2}} = 9 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358102 Two point charges exert on each other a force $F$ when they are placed $r$ distnce apart in air. If they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert double the force. The distance $R$ equals

1

\frac{r}{\sqrt{2 k}}

2

r k

3

\frac{r}{k}

4

r \sqrt{k}

Explanation:

In air medium the force is $F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
In a medium of dielectric constant $k$ , the force is
$F^{'} = \frac{q_{1} q_{2}}{4 π k ε R^{2}}$
As, $F^{'} = 2 F \Rightarrow \frac{q_{1} q_{2}}{4 π k ε_{0} R^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
$∴ R = \frac{r}{\sqrt{2 k}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358103 If point charges $Q_{1} = 2 \times 10^{- 7} C and Q_{2} = 3 \times 10^{- 7} C$ are at $30 cm$ separation, then find electrostatic force between them.

1

2 \times 10^{- 3} N

2

6 \times 10^{- 3} N

3

5 \times 10^{- 3} N

4

1 \times 10^{- 3} N

Explanation:

$∵$ Electrostatic force, $F = k \frac{Q_{1} Q_{2}}{r^{2}}$
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 7} \times 3 \times 10^{- 7}}{{(30 \times 10^{- 2})}^{2}}$
$= 6 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358100 A point charge $Q$ is fixed at the centre of an insulated disc of mass $M$ . The disc rests on a rough horizontal plane. Another charge having same value as point charge is fixed vertically above the centre of the disc at a height of $h$ . After the disc is displaced slightly in the horizontal direction (friction is sufficient to prevent slipping). If the period of oscillation of disc is found to be $2 π \frac{h}{Q} \sqrt{n π ε_{0} M h}$ ,
supporting img

1 2

2 6

3 8

4 10

Explanation:

Let the radius of the disc be $R$ . If the disc is displaced $x$ , the corresponding angular displacement $θ = x / R$ .
The restoring torque $τ$ about the point of contact of disc with ground,
$τ_{p} = (F \sin θ) R$
$(F \sin θ)$ $R = I α = [\frac{M R^{2}}{2} + M R^{2}] α (1)$
where $F = \frac{Q^{2}}{4 π ε_{0} (h^{2} + x^{2})} (2)$

supporting img

and $\sin θ = \frac{x}{\sqrt{h^{2} + x^{2}}} (3)$
From (1), (2) and (3), we get
$\frac{Q^{2} x R}{4 π ε_{0} {(h^{2} + x^{2})}^{3 / 2}} = [\frac{M R^{2}}{2} + M R^{2}] α$
$o r α = \frac{Q^{2} x}{6 π ε_{0} M R {(h^{2} + x^{2})}^{3 / 2}}$
$F o r x << h, α ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
$a n d α = \frac{a}{R} ≃ - \frac{Q^{2} x}{6 π ε_{0} M R h^{3}}$
Negative sign is being introduced because angular acceleration and angular displacement are opposite to each other.
Thus, $a = - \frac{Q^{2} x}{6 π ε_{0} M h^{3}}$
Hence, $ω = \sqrt{\frac{Q^{2}}{6 π ε_{0} M h^{3}}}$
or $T = 2 π \sqrt{\frac{6 π ε_{0} M h^{3}}{Q^{2}}} = 2 π \frac{h}{Q} \sqrt{6 π ε_{0} M h}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358101 Two equal point charges each of $3 μ C$ are separated by a certain distance in metres. If they are located at $(\hat{i} + \hat{j} + \hat{k})$ and $(2 \hat{i} + 3 \hat{j} + 3 \hat{k}),$ the electrostatic force between them is

1

9 \times 10^{- 2} N

2

9 \times 10^{- 3} N

3

10^{- 3} N

4

9 \times 10^{3} N

Explanation:

Here, $q_{1} = 3 \times 10^{- 6} C, q_{2} = 3 \times 10^{- 6} C$
$r_{1} = \hat{i} + \hat{j} + \hat{k}, r_{2} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$
$∴ \vec{r} = \vec{r_{2}} - \vec{r_{1}}$
$= (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}$
The distance between the two charges is
$| \vec{r_{1}} | = \sqrt{{(1)}^{2} + {(2)}^{2} + {(2)}^{2}} = 3$
According to
coulomb’s
law,
$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6} \times 3 \times 10^{- 6}}{{(3)}^{2}} = 9 \times 10^{- 3} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358102 Two point charges exert on each other a force $F$ when they are placed $r$ distnce apart in air. If they are placed $R$ distance apart in a medium of dielectric constant $K$ , they exert double the force. The distance $R$ equals

1

\frac{r}{\sqrt{2 k}}

2

r k

3

\frac{r}{k}

4

r \sqrt{k}

Explanation:

In air medium the force is $F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
In a medium of dielectric constant $k$ , the force is
$F^{'} = \frac{q_{1} q_{2}}{4 π k ε R^{2}}$
As, $F^{'} = 2 F \Rightarrow \frac{q_{1} q_{2}}{4 π k ε_{0} R^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
$∴ R = \frac{r}{\sqrt{2 k}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358103 If point charges $Q_{1} = 2 \times 10^{- 7} C and Q_{2} = 3 \times 10^{- 7} C$ are at $30 cm$ separation, then find electrostatic force between them.

1

2 \times 10^{- 3} N

2

6 \times 10^{- 3} N

3

5 \times 10^{- 3} N

4

1 \times 10^{- 3} N

Explanation:

$∵$ Electrostatic force, $F = k \frac{Q_{1} Q_{2}}{r^{2}}$
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 7} \times 3 \times 10^{- 7}}{{(30 \times 10^{- 2})}^{2}}$
$= 6 \times 10^{- 3} N$

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