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PHXII01:ELECTRIC CHARGES AND FIELDS

358104 Two identical metal spheres possesses $+ 60 C$ and $- 20 C$ of charges. They are brought in contact and the then separated by $10 c m$ . The force between them is

1

36 \times 10^{13} N

2

36 \times 10^{14} N

3

36 \times 10^{12} N

4

3.6 \times 10^{12} N

Explanation:

When two charges are brought in contact
$q_{1} + q_{2} = 60 - 20 = 40 C$
The two spheres will get equal share, i.e., $20 C$ each.
The force between the charges is
$F = \frac{1}{4 π ε_{0}} \frac{(20) (20)}{{(0.1)}^{2}}$
$= \frac{9 \times 10^{9} \times 400}{10^{- 2}} = 36 \times 10^{13} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358105 Charges on two spheres are $+ 10 μ C$ and $- 5 μ C$ respectively. They experience a force $F$ . If each of them is given an additional charge $+ 2 μ C$ then new force between them keeping the same distance is

1

18 F

2

\frac{F}{25}

3

\frac{18 F}{25}

4

\frac{25}{18} F

Explanation:

When there is no additional charge
$\begin{aligned} F = \frac{K (+ 10) (5)}{R^{2}} \\ F = \frac{- 50 K}{R^{2}} \end{aligned}$
when $+ 2 μ C$ charge is added to both spheres.
Charge on sphere $1^{s t} = 10 μ C + 2 μ C = 12 μ C$
Charge on sphere $2^{nd} = - 5 μ C + 2 μ C = - 3 μ C$
New force will be $F^{'} = \frac{K (+ 12) (- 3)}{R^{2}} = - \frac{36 K}{R^{2}}$
$∴ \frac{F^{'}}{F} = - \frac{36 K}{R^{2}} \times \frac{R^{2}}{- 50 K} \Rightarrow \frac{F^{'}}{F} = \frac{18}{25}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358106 Force between $A$ and $B$ is $F$ as shown in the figure. If $50 %$ charge $A$ of is transferred to $B$ , then force between $A$ and $B$ is
supporting img

1

\frac{F}{3}

2

4 F

3

\frac{5 F}{2}

4

2 F

Explanation:

For the given combination of charges
supporting img
$\Rightarrow F = \frac{k (8 Q) (Q)}{r^{2}} = \frac{8 k Q^{2}}{r^{2}}$

$\Rightarrow F^{'} = \frac{k (4 Q) (5 Q)}{r^{2}} = \frac{5 F}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358107 Force between two identical charges placed at a distance of $r$ in vacuum is $F$ . Now a slab of dielectric of dielectric contrant 4 is inserted between these two charges. If the thickness of the slab is $r / 2$ , then the force between the charges will become

1

F

2

\frac{3}{5} F

3

\frac{4}{9} F

4

\frac{F}{2}

Explanation:

In vacuum, $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} (1)$
Suppose, force between the charges is same when charges are $r^{'}$ distance apart in dielectric. $∴ F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{k r^{' 2}} (2)$
From (1) and (2) $k r^{' 2} = r^{2}$ or, $r = \sqrt{k r^{'}}$
In the given situation, force between the charges would be
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + \sqrt{4} \frac{r}{2})}^{2}} = \frac{4}{9} \frac{q^{2}}{4 π ε_{0} r^{2}} = \frac{4 F}{9}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358104 Two identical metal spheres possesses $+ 60 C$ and $- 20 C$ of charges. They are brought in contact and the then separated by $10 c m$ . The force between them is

1

36 \times 10^{13} N

2

36 \times 10^{14} N

3

36 \times 10^{12} N

4

3.6 \times 10^{12} N

Explanation:

When two charges are brought in contact
$q_{1} + q_{2} = 60 - 20 = 40 C$
The two spheres will get equal share, i.e., $20 C$ each.
The force between the charges is
$F = \frac{1}{4 π ε_{0}} \frac{(20) (20)}{{(0.1)}^{2}}$
$= \frac{9 \times 10^{9} \times 400}{10^{- 2}} = 36 \times 10^{13} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358105 Charges on two spheres are $+ 10 μ C$ and $- 5 μ C$ respectively. They experience a force $F$ . If each of them is given an additional charge $+ 2 μ C$ then new force between them keeping the same distance is

1

18 F

2

\frac{F}{25}

3

\frac{18 F}{25}

4

\frac{25}{18} F

Explanation:

When there is no additional charge
$\begin{aligned} F = \frac{K (+ 10) (5)}{R^{2}} \\ F = \frac{- 50 K}{R^{2}} \end{aligned}$
when $+ 2 μ C$ charge is added to both spheres.
Charge on sphere $1^{s t} = 10 μ C + 2 μ C = 12 μ C$
Charge on sphere $2^{nd} = - 5 μ C + 2 μ C = - 3 μ C$
New force will be $F^{'} = \frac{K (+ 12) (- 3)}{R^{2}} = - \frac{36 K}{R^{2}}$
$∴ \frac{F^{'}}{F} = - \frac{36 K}{R^{2}} \times \frac{R^{2}}{- 50 K} \Rightarrow \frac{F^{'}}{F} = \frac{18}{25}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358106 Force between $A$ and $B$ is $F$ as shown in the figure. If $50 %$ charge $A$ of is transferred to $B$ , then force between $A$ and $B$ is
supporting img

1

\frac{F}{3}

2

4 F

3

\frac{5 F}{2}

4

2 F

Explanation:

For the given combination of charges
supporting img
$\Rightarrow F = \frac{k (8 Q) (Q)}{r^{2}} = \frac{8 k Q^{2}}{r^{2}}$

$\Rightarrow F^{'} = \frac{k (4 Q) (5 Q)}{r^{2}} = \frac{5 F}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358107 Force between two identical charges placed at a distance of $r$ in vacuum is $F$ . Now a slab of dielectric of dielectric contrant 4 is inserted between these two charges. If the thickness of the slab is $r / 2$ , then the force between the charges will become

1

F

2

\frac{3}{5} F

3

\frac{4}{9} F

4

\frac{F}{2}

Explanation:

In vacuum, $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} (1)$
Suppose, force between the charges is same when charges are $r^{'}$ distance apart in dielectric. $∴ F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{k r^{' 2}} (2)$
From (1) and (2) $k r^{' 2} = r^{2}$ or, $r = \sqrt{k r^{'}}$
In the given situation, force between the charges would be
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + \sqrt{4} \frac{r}{2})}^{2}} = \frac{4}{9} \frac{q^{2}}{4 π ε_{0} r^{2}} = \frac{4 F}{9}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358104 Two identical metal spheres possesses $+ 60 C$ and $- 20 C$ of charges. They are brought in contact and the then separated by $10 c m$ . The force between them is

1

36 \times 10^{13} N

2

36 \times 10^{14} N

3

36 \times 10^{12} N

4

3.6 \times 10^{12} N

Explanation:

When two charges are brought in contact
$q_{1} + q_{2} = 60 - 20 = 40 C$
The two spheres will get equal share, i.e., $20 C$ each.
The force between the charges is
$F = \frac{1}{4 π ε_{0}} \frac{(20) (20)}{{(0.1)}^{2}}$
$= \frac{9 \times 10^{9} \times 400}{10^{- 2}} = 36 \times 10^{13} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358105 Charges on two spheres are $+ 10 μ C$ and $- 5 μ C$ respectively. They experience a force $F$ . If each of them is given an additional charge $+ 2 μ C$ then new force between them keeping the same distance is

1

18 F

2

\frac{F}{25}

3

\frac{18 F}{25}

4

\frac{25}{18} F

Explanation:

When there is no additional charge
$\begin{aligned} F = \frac{K (+ 10) (5)}{R^{2}} \\ F = \frac{- 50 K}{R^{2}} \end{aligned}$
when $+ 2 μ C$ charge is added to both spheres.
Charge on sphere $1^{s t} = 10 μ C + 2 μ C = 12 μ C$
Charge on sphere $2^{nd} = - 5 μ C + 2 μ C = - 3 μ C$
New force will be $F^{'} = \frac{K (+ 12) (- 3)}{R^{2}} = - \frac{36 K}{R^{2}}$
$∴ \frac{F^{'}}{F} = - \frac{36 K}{R^{2}} \times \frac{R^{2}}{- 50 K} \Rightarrow \frac{F^{'}}{F} = \frac{18}{25}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358106 Force between $A$ and $B$ is $F$ as shown in the figure. If $50 %$ charge $A$ of is transferred to $B$ , then force between $A$ and $B$ is
supporting img

1

\frac{F}{3}

2

4 F

3

\frac{5 F}{2}

4

2 F

Explanation:

For the given combination of charges
supporting img
$\Rightarrow F = \frac{k (8 Q) (Q)}{r^{2}} = \frac{8 k Q^{2}}{r^{2}}$

$\Rightarrow F^{'} = \frac{k (4 Q) (5 Q)}{r^{2}} = \frac{5 F}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358107 Force between two identical charges placed at a distance of $r$ in vacuum is $F$ . Now a slab of dielectric of dielectric contrant 4 is inserted between these two charges. If the thickness of the slab is $r / 2$ , then the force between the charges will become

1

F

2

\frac{3}{5} F

3

\frac{4}{9} F

4

\frac{F}{2}

Explanation:

In vacuum, $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} (1)$
Suppose, force between the charges is same when charges are $r^{'}$ distance apart in dielectric. $∴ F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{k r^{' 2}} (2)$
From (1) and (2) $k r^{' 2} = r^{2}$ or, $r = \sqrt{k r^{'}}$
In the given situation, force between the charges would be
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + \sqrt{4} \frac{r}{2})}^{2}} = \frac{4}{9} \frac{q^{2}}{4 π ε_{0} r^{2}} = \frac{4 F}{9}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358104 Two identical metal spheres possesses $+ 60 C$ and $- 20 C$ of charges. They are brought in contact and the then separated by $10 c m$ . The force between them is

1

36 \times 10^{13} N

2

36 \times 10^{14} N

3

36 \times 10^{12} N

4

3.6 \times 10^{12} N

Explanation:

When two charges are brought in contact
$q_{1} + q_{2} = 60 - 20 = 40 C$
The two spheres will get equal share, i.e., $20 C$ each.
The force between the charges is
$F = \frac{1}{4 π ε_{0}} \frac{(20) (20)}{{(0.1)}^{2}}$
$= \frac{9 \times 10^{9} \times 400}{10^{- 2}} = 36 \times 10^{13} N$

PHXII01:ELECTRIC CHARGES AND FIELDS

358105 Charges on two spheres are $+ 10 μ C$ and $- 5 μ C$ respectively. They experience a force $F$ . If each of them is given an additional charge $+ 2 μ C$ then new force between them keeping the same distance is

1

18 F

2

\frac{F}{25}

3

\frac{18 F}{25}

4

\frac{25}{18} F

Explanation:

When there is no additional charge
$\begin{aligned} F = \frac{K (+ 10) (5)}{R^{2}} \\ F = \frac{- 50 K}{R^{2}} \end{aligned}$
when $+ 2 μ C$ charge is added to both spheres.
Charge on sphere $1^{s t} = 10 μ C + 2 μ C = 12 μ C$
Charge on sphere $2^{nd} = - 5 μ C + 2 μ C = - 3 μ C$
New force will be $F^{'} = \frac{K (+ 12) (- 3)}{R^{2}} = - \frac{36 K}{R^{2}}$
$∴ \frac{F^{'}}{F} = - \frac{36 K}{R^{2}} \times \frac{R^{2}}{- 50 K} \Rightarrow \frac{F^{'}}{F} = \frac{18}{25}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358106 Force between $A$ and $B$ is $F$ as shown in the figure. If $50 %$ charge $A$ of is transferred to $B$ , then force between $A$ and $B$ is
supporting img

1

\frac{F}{3}

2

4 F

3

\frac{5 F}{2}

4

2 F

Explanation:

For the given combination of charges
supporting img
$\Rightarrow F = \frac{k (8 Q) (Q)}{r^{2}} = \frac{8 k Q^{2}}{r^{2}}$

$\Rightarrow F^{'} = \frac{k (4 Q) (5 Q)}{r^{2}} = \frac{5 F}{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358107 Force between two identical charges placed at a distance of $r$ in vacuum is $F$ . Now a slab of dielectric of dielectric contrant 4 is inserted between these two charges. If the thickness of the slab is $r / 2$ , then the force between the charges will become

1

F

2

\frac{3}{5} F

3

\frac{4}{9} F

4

\frac{F}{2}

Explanation:

In vacuum, $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}} (1)$
Suppose, force between the charges is same when charges are $r^{'}$ distance apart in dielectric. $∴ F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{k r^{' 2}} (2)$
From (1) and (2) $k r^{' 2} = r^{2}$ or, $r = \sqrt{k r^{'}}$
In the given situation, force between the charges would be
$F^{'} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{{(\frac{r}{2} + \sqrt{4} \frac{r}{2})}^{2}} = \frac{4}{9} \frac{q^{2}}{4 π ε_{0} r^{2}} = \frac{4 F}{9}$

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