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PHXII01:ELECTRIC CHARGES AND FIELDS

358108 A total charge $Q$ is broken in two parts $Q_{1}$ and $Q_{2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur. When

1

Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}

2

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}

3

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{3 Q}{4}

4

Q_{1} = \frac{Q}{2}, Q_{2} = Q - \frac{Q}{2}

Explanation:

$Q_{1} + Q_{2} = Q (1)$
and $F = K \frac{Q_{1} Q_{2}}{2} (2)$
From (1) and (2) $F = \frac{K Q_{1} (Q - Q_{1})}{r^{2}}$
For $F$ to be maximum $\frac{d F}{d Q_{1}} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358109 Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$ , then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

1

\frac{F}{4}

2

\frac{3 F}{4}

3

\frac{F}{8}

4

\frac{3 F}{8}

Explanation:

Let the spherical conductors $B$ and $C$ have same charge as $q$ .
The electric force between them is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
where, $r$ being the distance between them.
When third uncharged conductor $A$ is brought in contact with $B$ , then charge on each conductor
$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2}$
$= \frac{0 + q}{2} = \frac{q}{2}$
When this conductor $A$ is now brought in contact with $C$ , then charge on each conductor
$q_{A} = q_{C} = \frac{q_{A} + q_{C}}{2}$
$= \frac{(q / 2) + q}{2} = \frac{3 q}{4}$
Hence, electric force acting between $B$ and $C$ is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q_{B} q_{C}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{(q / 2) (3 q / 4)}{r^{2}}$
$= \frac{3}{8} [\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}] = \frac{3 F}{8}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358110 An isolated charge $q_{1}$ of mass $m$ is suspended freely by a thread of length $l$ . Another charge $q_{2}$ is brought near it $(r >> l)$ .When $q_{1}$ is in equilibrium, tension in thread will be
supporting img

1

m g

2

> m g

3

< m g

4 None of these

Explanation:

Initial tension: $T_{1} = m g$
Final tension: $T_{2} c o s θ = m g$
or $T_{2} = \frac{m g}{\cos θ}$
Here we have assumed $θ$ to be small so that $F$ is almost horizontal.
$T_{2} > m g .$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358111 The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$ . Half the gap between the two charges is filled by a dielectic slab of dielectric constant 16.Then the new force of repulsion between those two charges becomes

1

\frac{F}{3}

2

\frac{2 F}{3}

3

\frac{4 F}{3}

4

\frac{4 F}{25}

Explanation:

Let $q$ be each charge,
According to Coloumb's law, the force of repulsion between the two charges when kept in air is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
When half of the gap between the two charges is filled with the dielectric slab of dielectric constant $(K = 16)$ , the new force of repulsion between them becomes
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{((r - t) + t \sqrt{K)^{2}}}$
(where $t$ is the thickness of the slab)
$= \frac{1}{4 π ε_{0}} \frac{q^{2}}{{[(r - \frac{r}{2}) + \frac{r}{2} \sqrt{16}]}^{2}}$
$(a s t = \frac{r}{2}, K = 16)$
$\frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + 2 r)}^{2}} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{5}{2} r)}^{2}} = \frac{4 F}{25}$

KCET - 2018

PHXII01:ELECTRIC CHARGES AND FIELDS

358108 A total charge $Q$ is broken in two parts $Q_{1}$ and $Q_{2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur. When

1

Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}

2

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}

3

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{3 Q}{4}

4

Q_{1} = \frac{Q}{2}, Q_{2} = Q - \frac{Q}{2}

Explanation:

$Q_{1} + Q_{2} = Q (1)$
and $F = K \frac{Q_{1} Q_{2}}{2} (2)$
From (1) and (2) $F = \frac{K Q_{1} (Q - Q_{1})}{r^{2}}$
For $F$ to be maximum $\frac{d F}{d Q_{1}} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358109 Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$ , then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

1

\frac{F}{4}

2

\frac{3 F}{4}

3

\frac{F}{8}

4

\frac{3 F}{8}

Explanation:

Let the spherical conductors $B$ and $C$ have same charge as $q$ .
The electric force between them is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
where, $r$ being the distance between them.
When third uncharged conductor $A$ is brought in contact with $B$ , then charge on each conductor
$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2}$
$= \frac{0 + q}{2} = \frac{q}{2}$
When this conductor $A$ is now brought in contact with $C$ , then charge on each conductor
$q_{A} = q_{C} = \frac{q_{A} + q_{C}}{2}$
$= \frac{(q / 2) + q}{2} = \frac{3 q}{4}$
Hence, electric force acting between $B$ and $C$ is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q_{B} q_{C}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{(q / 2) (3 q / 4)}{r^{2}}$
$= \frac{3}{8} [\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}] = \frac{3 F}{8}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358110 An isolated charge $q_{1}$ of mass $m$ is suspended freely by a thread of length $l$ . Another charge $q_{2}$ is brought near it $(r >> l)$ .When $q_{1}$ is in equilibrium, tension in thread will be
supporting img

1

m g

2

> m g

3

< m g

4 None of these

Explanation:

Initial tension: $T_{1} = m g$
Final tension: $T_{2} c o s θ = m g$
or $T_{2} = \frac{m g}{\cos θ}$
Here we have assumed $θ$ to be small so that $F$ is almost horizontal.
$T_{2} > m g .$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358111 The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$ . Half the gap between the two charges is filled by a dielectic slab of dielectric constant 16.Then the new force of repulsion between those two charges becomes

1

\frac{F}{3}

2

\frac{2 F}{3}

3

\frac{4 F}{3}

4

\frac{4 F}{25}

Explanation:

Let $q$ be each charge,
According to Coloumb's law, the force of repulsion between the two charges when kept in air is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
When half of the gap between the two charges is filled with the dielectric slab of dielectric constant $(K = 16)$ , the new force of repulsion between them becomes
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{((r - t) + t \sqrt{K)^{2}}}$
(where $t$ is the thickness of the slab)
$= \frac{1}{4 π ε_{0}} \frac{q^{2}}{{[(r - \frac{r}{2}) + \frac{r}{2} \sqrt{16}]}^{2}}$
$(a s t = \frac{r}{2}, K = 16)$
$\frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + 2 r)}^{2}} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{5}{2} r)}^{2}} = \frac{4 F}{25}$

KCET - 2018

PHXII01:ELECTRIC CHARGES AND FIELDS

358108 A total charge $Q$ is broken in two parts $Q_{1}$ and $Q_{2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur. When

1

Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}

2

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}

3

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{3 Q}{4}

4

Q_{1} = \frac{Q}{2}, Q_{2} = Q - \frac{Q}{2}

Explanation:

$Q_{1} + Q_{2} = Q (1)$
and $F = K \frac{Q_{1} Q_{2}}{2} (2)$
From (1) and (2) $F = \frac{K Q_{1} (Q - Q_{1})}{r^{2}}$
For $F$ to be maximum $\frac{d F}{d Q_{1}} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358109 Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$ , then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

1

\frac{F}{4}

2

\frac{3 F}{4}

3

\frac{F}{8}

4

\frac{3 F}{8}

Explanation:

Let the spherical conductors $B$ and $C$ have same charge as $q$ .
The electric force between them is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
where, $r$ being the distance between them.
When third uncharged conductor $A$ is brought in contact with $B$ , then charge on each conductor
$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2}$
$= \frac{0 + q}{2} = \frac{q}{2}$
When this conductor $A$ is now brought in contact with $C$ , then charge on each conductor
$q_{A} = q_{C} = \frac{q_{A} + q_{C}}{2}$
$= \frac{(q / 2) + q}{2} = \frac{3 q}{4}$
Hence, electric force acting between $B$ and $C$ is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q_{B} q_{C}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{(q / 2) (3 q / 4)}{r^{2}}$
$= \frac{3}{8} [\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}] = \frac{3 F}{8}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358110 An isolated charge $q_{1}$ of mass $m$ is suspended freely by a thread of length $l$ . Another charge $q_{2}$ is brought near it $(r >> l)$ .When $q_{1}$ is in equilibrium, tension in thread will be
supporting img

1

m g

2

> m g

3

< m g

4 None of these

Explanation:

Initial tension: $T_{1} = m g$
Final tension: $T_{2} c o s θ = m g$
or $T_{2} = \frac{m g}{\cos θ}$
Here we have assumed $θ$ to be small so that $F$ is almost horizontal.
$T_{2} > m g .$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358111 The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$ . Half the gap between the two charges is filled by a dielectic slab of dielectric constant 16.Then the new force of repulsion between those two charges becomes

1

\frac{F}{3}

2

\frac{2 F}{3}

3

\frac{4 F}{3}

4

\frac{4 F}{25}

Explanation:

Let $q$ be each charge,
According to Coloumb's law, the force of repulsion between the two charges when kept in air is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
When half of the gap between the two charges is filled with the dielectric slab of dielectric constant $(K = 16)$ , the new force of repulsion between them becomes
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{((r - t) + t \sqrt{K)^{2}}}$
(where $t$ is the thickness of the slab)
$= \frac{1}{4 π ε_{0}} \frac{q^{2}}{{[(r - \frac{r}{2}) + \frac{r}{2} \sqrt{16}]}^{2}}$
$(a s t = \frac{r}{2}, K = 16)$
$\frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + 2 r)}^{2}} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{5}{2} r)}^{2}} = \frac{4 F}{25}$

KCET - 2018

PHXII01:ELECTRIC CHARGES AND FIELDS

358108 A total charge $Q$ is broken in two parts $Q_{1}$ and $Q_{2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur. When

1

Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}

2

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}

3

Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{3 Q}{4}

4

Q_{1} = \frac{Q}{2}, Q_{2} = Q - \frac{Q}{2}

Explanation:

$Q_{1} + Q_{2} = Q (1)$
and $F = K \frac{Q_{1} Q_{2}}{2} (2)$
From (1) and (2) $F = \frac{K Q_{1} (Q - Q_{1})}{r^{2}}$
For $F$ to be maximum $\frac{d F}{d Q_{1}} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358109 Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$ , then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

1

\frac{F}{4}

2

\frac{3 F}{4}

3

\frac{F}{8}

4

\frac{3 F}{8}

Explanation:

Let the spherical conductors $B$ and $C$ have same charge as $q$ .
The electric force between them is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
where, $r$ being the distance between them.
When third uncharged conductor $A$ is brought in contact with $B$ , then charge on each conductor
$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2}$
$= \frac{0 + q}{2} = \frac{q}{2}$
When this conductor $A$ is now brought in contact with $C$ , then charge on each conductor
$q_{A} = q_{C} = \frac{q_{A} + q_{C}}{2}$
$= \frac{(q / 2) + q}{2} = \frac{3 q}{4}$
Hence, electric force acting between $B$ and $C$ is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q_{B} q_{C}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{(q / 2) (3 q / 4)}{r^{2}}$
$= \frac{3}{8} [\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}] = \frac{3 F}{8}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358110 An isolated charge $q_{1}$ of mass $m$ is suspended freely by a thread of length $l$ . Another charge $q_{2}$ is brought near it $(r >> l)$ .When $q_{1}$ is in equilibrium, tension in thread will be
supporting img

1

m g

2

> m g

3

< m g

4 None of these

Explanation:

Initial tension: $T_{1} = m g$
Final tension: $T_{2} c o s θ = m g$
or $T_{2} = \frac{m g}{\cos θ}$
Here we have assumed $θ$ to be small so that $F$ is almost horizontal.
$T_{2} > m g .$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358111 The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$ . Half the gap between the two charges is filled by a dielectic slab of dielectric constant 16.Then the new force of repulsion between those two charges becomes

1

\frac{F}{3}

2

\frac{2 F}{3}

3

\frac{4 F}{3}

4

\frac{4 F}{25}

Explanation:

Let $q$ be each charge,
According to Coloumb's law, the force of repulsion between the two charges when kept in air is
$F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
When half of the gap between the two charges is filled with the dielectric slab of dielectric constant $(K = 16)$ , the new force of repulsion between them becomes
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{((r - t) + t \sqrt{K)^{2}}}$
(where $t$ is the thickness of the slab)
$= \frac{1}{4 π ε_{0}} \frac{q^{2}}{{[(r - \frac{r}{2}) + \frac{r}{2} \sqrt{16}]}^{2}}$
$(a s t = \frac{r}{2}, K = 16)$
$\frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + 2 r)}^{2}} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{5}{2} r)}^{2}} = \frac{4 F}{25}$

KCET - 2018

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