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PHXII01:ELECTRIC CHARGES AND FIELDS

358052 Two small spheres of masses $M_{1} a n d M_{2}$ are suspended by weightless insulating threads of lengths $L_{1} a n d L_{2}$ . The spheres carry charges $Q_{1} a n d Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $θ_{1} a n d θ_{2}$ as shown in figure. Which one of the following conditions is essential, if $θ_{1} = θ_{2} ?$
supporting img

1

M_{1} \neq M_{2} b u t Q_{1} = Q_{2}

2

M_{1} = M_{2}

3

Q_{1} = Q_{2}

4

L_{1} = L_{2}

Explanation:

The three forces acting on each sphere are;
$i) T e n s i o n i i) W e i g h t$
$i i i) E l e c t r o s t a t i c f o r c e o f r e p u l s i o n$
For sphere 1
In equilibrium, from figure,
$T_{1} \cos θ_{1} = M_{1} g; T_{1} \sin θ_{1} = F_{1}$
$∴ \tan θ_{1} = \frac{F_{1}}{M_{1} g}$
For sphere 2
In equilibrium, from figure
$T_{2} \cos θ_{1} = M_{2} g; T_{2} \sin θ_{2} = F_{2}$
$∴ \tan θ_{2} = \frac{F_{2}}{M_{2} g}$
Force of repulsion between two charges are same.
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2} =$ only if
$\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
supporting img

KCET - 2010

PHXII01:ELECTRIC CHARGES AND FIELDS

358053 Assertion :
A charge is lying at the centre of the line joining two similar charges each which are fixed. The system will be in equilibrium if that charge is one fourth of the similar charges.
Reason :
For charge to be in equilibrium, sum of the forces on charge due to rest of the two charges must be zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

If centre - charge is not present, original charges (given to be similar, say: $+ Q$ and $+ Q$ ) will have coulombic repulsion force between them.
supporting img
In order to nullify it and keep system in equilibrium, the centre - charge that to be placed shall be of different sign (here $q$ shall be negative).
$\frac{1}{4 π ϵ_{0}} [\frac{Q Q}{d^{2}}] = \frac{1}{4 π ϵ_{0}} [\frac{Q (- q)}{(d / 2)^{2}}]$
$\Rightarrow | q | = | \frac{Q}{4} | \Rightarrow q = (- Q / 4)$
For the charges to be in equilibrium, the vector sum of the forces due to the rest of the two charges must be zero. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358054 Two small spheres each of mass $m$ and with a charge $q$ lie inside a nonconducting smooth hemispherical bowl of radius $R$ . Find $q$ if the equilibrium separation between the two charges is $d$ .

1

\sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}

2

\sqrt{\frac{m g π ε_{0} d^{3}}{\sqrt{R^{2} - d^{2}}}}

3

\sqrt{\frac{m g ε_{0} d^{3}}{\sqrt{R^{2} + d^{2}}}}

4

\sqrt{m g ε_{0} d^{3}}

Explanation:

The two charges are shown in the figure.
By considering the equilibrium of any charge
$N \cos θ = \frac{q^{2}}{4 π ε_{0} d^{2}} (1)$
$N \sin θ = m g (2)$
From both the equations
$Tan θ = \frac{m g 4 π ε_{0} d^{2}}{q^{2}} = \frac{\sqrt{R^{2} - \frac{d^{2}}{4}}}{d / 2}$ $q = \sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358055 A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to

1

- \frac{Q}{2}

2

- \frac{Q}{4}

3

+ \frac{Q}{4}

4

\frac{Q}{2}

Explanation:

The central charge is in the equilibrium. If the system has to be in equilibrium then the net force on any of the charge $Q$ should be zero.
$\frac{1}{4 π ε_{0}} \frac{Q Q}{l^{2}} + \frac{1}{4 π ε_{0}} \frac{q Q}{{(\frac{l}{2})}^{2}} = 0$
$\Rightarrow q = \frac{- Q}{4}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358056 Two positively charged spheres of masses, $m_{1} a n d m_{2}$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$ . Charges on the two spheres are $q_{1} a n d q_{2}$ ,respectively. At equilibrium both strings make the same angle $θ$ with the vertical.Then

1

q_{1} m_{1} = q_{2} m_{2}

2

m_{1} = m_{2}

3

m_{1} = m_{2} \sin θ

4

q_{2} m_{1} = q_{1} m_{2}

Explanation:

The forces on the two masses are shown in the figure. Let $θ$ is the angle made by the strings with vertical
supporting img
For equilibrium of $m_{1}$
$T_{1} \cos θ = m_{1} g$
$T_{1} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (1)$
For equilibrium of $m_{2}$
$T_{2} \cos θ = m_{2} g$
$T_{2} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (2)$
From (1) & (2)
$m_{1} = m_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358052 Two small spheres of masses $M_{1} a n d M_{2}$ are suspended by weightless insulating threads of lengths $L_{1} a n d L_{2}$ . The spheres carry charges $Q_{1} a n d Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $θ_{1} a n d θ_{2}$ as shown in figure. Which one of the following conditions is essential, if $θ_{1} = θ_{2} ?$
supporting img

1

M_{1} \neq M_{2} b u t Q_{1} = Q_{2}

2

M_{1} = M_{2}

3

Q_{1} = Q_{2}

4

L_{1} = L_{2}

Explanation:

The three forces acting on each sphere are;
$i) T e n s i o n i i) W e i g h t$
$i i i) E l e c t r o s t a t i c f o r c e o f r e p u l s i o n$
For sphere 1
In equilibrium, from figure,
$T_{1} \cos θ_{1} = M_{1} g; T_{1} \sin θ_{1} = F_{1}$
$∴ \tan θ_{1} = \frac{F_{1}}{M_{1} g}$
For sphere 2
In equilibrium, from figure
$T_{2} \cos θ_{1} = M_{2} g; T_{2} \sin θ_{2} = F_{2}$
$∴ \tan θ_{2} = \frac{F_{2}}{M_{2} g}$
Force of repulsion between two charges are same.
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2} =$ only if
$\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
supporting img

KCET - 2010

PHXII01:ELECTRIC CHARGES AND FIELDS

358053 Assertion :
A charge is lying at the centre of the line joining two similar charges each which are fixed. The system will be in equilibrium if that charge is one fourth of the similar charges.
Reason :
For charge to be in equilibrium, sum of the forces on charge due to rest of the two charges must be zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

If centre - charge is not present, original charges (given to be similar, say: $+ Q$ and $+ Q$ ) will have coulombic repulsion force between them.
supporting img
In order to nullify it and keep system in equilibrium, the centre - charge that to be placed shall be of different sign (here $q$ shall be negative).
$\frac{1}{4 π ϵ_{0}} [\frac{Q Q}{d^{2}}] = \frac{1}{4 π ϵ_{0}} [\frac{Q (- q)}{(d / 2)^{2}}]$
$\Rightarrow | q | = | \frac{Q}{4} | \Rightarrow q = (- Q / 4)$
For the charges to be in equilibrium, the vector sum of the forces due to the rest of the two charges must be zero. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358054 Two small spheres each of mass $m$ and with a charge $q$ lie inside a nonconducting smooth hemispherical bowl of radius $R$ . Find $q$ if the equilibrium separation between the two charges is $d$ .

1

\sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}

2

\sqrt{\frac{m g π ε_{0} d^{3}}{\sqrt{R^{2} - d^{2}}}}

3

\sqrt{\frac{m g ε_{0} d^{3}}{\sqrt{R^{2} + d^{2}}}}

4

\sqrt{m g ε_{0} d^{3}}

Explanation:

The two charges are shown in the figure.
By considering the equilibrium of any charge
$N \cos θ = \frac{q^{2}}{4 π ε_{0} d^{2}} (1)$
$N \sin θ = m g (2)$
From both the equations
$Tan θ = \frac{m g 4 π ε_{0} d^{2}}{q^{2}} = \frac{\sqrt{R^{2} - \frac{d^{2}}{4}}}{d / 2}$ $q = \sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358055 A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to

1

- \frac{Q}{2}

2

- \frac{Q}{4}

3

+ \frac{Q}{4}

4

\frac{Q}{2}

Explanation:

The central charge is in the equilibrium. If the system has to be in equilibrium then the net force on any of the charge $Q$ should be zero.
$\frac{1}{4 π ε_{0}} \frac{Q Q}{l^{2}} + \frac{1}{4 π ε_{0}} \frac{q Q}{{(\frac{l}{2})}^{2}} = 0$
$\Rightarrow q = \frac{- Q}{4}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358056 Two positively charged spheres of masses, $m_{1} a n d m_{2}$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$ . Charges on the two spheres are $q_{1} a n d q_{2}$ ,respectively. At equilibrium both strings make the same angle $θ$ with the vertical.Then

1

q_{1} m_{1} = q_{2} m_{2}

2

m_{1} = m_{2}

3

m_{1} = m_{2} \sin θ

4

q_{2} m_{1} = q_{1} m_{2}

Explanation:

The forces on the two masses are shown in the figure. Let $θ$ is the angle made by the strings with vertical
supporting img
For equilibrium of $m_{1}$
$T_{1} \cos θ = m_{1} g$
$T_{1} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (1)$
For equilibrium of $m_{2}$
$T_{2} \cos θ = m_{2} g$
$T_{2} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (2)$
From (1) & (2)
$m_{1} = m_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358052 Two small spheres of masses $M_{1} a n d M_{2}$ are suspended by weightless insulating threads of lengths $L_{1} a n d L_{2}$ . The spheres carry charges $Q_{1} a n d Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $θ_{1} a n d θ_{2}$ as shown in figure. Which one of the following conditions is essential, if $θ_{1} = θ_{2} ?$
supporting img

1

M_{1} \neq M_{2} b u t Q_{1} = Q_{2}

2

M_{1} = M_{2}

3

Q_{1} = Q_{2}

4

L_{1} = L_{2}

Explanation:

The three forces acting on each sphere are;
$i) T e n s i o n i i) W e i g h t$
$i i i) E l e c t r o s t a t i c f o r c e o f r e p u l s i o n$
For sphere 1
In equilibrium, from figure,
$T_{1} \cos θ_{1} = M_{1} g; T_{1} \sin θ_{1} = F_{1}$
$∴ \tan θ_{1} = \frac{F_{1}}{M_{1} g}$
For sphere 2
In equilibrium, from figure
$T_{2} \cos θ_{1} = M_{2} g; T_{2} \sin θ_{2} = F_{2}$
$∴ \tan θ_{2} = \frac{F_{2}}{M_{2} g}$
Force of repulsion between two charges are same.
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2} =$ only if
$\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
supporting img

KCET - 2010

PHXII01:ELECTRIC CHARGES AND FIELDS

358053 Assertion :
A charge is lying at the centre of the line joining two similar charges each which are fixed. The system will be in equilibrium if that charge is one fourth of the similar charges.
Reason :
For charge to be in equilibrium, sum of the forces on charge due to rest of the two charges must be zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

If centre - charge is not present, original charges (given to be similar, say: $+ Q$ and $+ Q$ ) will have coulombic repulsion force between them.
supporting img
In order to nullify it and keep system in equilibrium, the centre - charge that to be placed shall be of different sign (here $q$ shall be negative).
$\frac{1}{4 π ϵ_{0}} [\frac{Q Q}{d^{2}}] = \frac{1}{4 π ϵ_{0}} [\frac{Q (- q)}{(d / 2)^{2}}]$
$\Rightarrow | q | = | \frac{Q}{4} | \Rightarrow q = (- Q / 4)$
For the charges to be in equilibrium, the vector sum of the forces due to the rest of the two charges must be zero. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358054 Two small spheres each of mass $m$ and with a charge $q$ lie inside a nonconducting smooth hemispherical bowl of radius $R$ . Find $q$ if the equilibrium separation between the two charges is $d$ .

1

\sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}

2

\sqrt{\frac{m g π ε_{0} d^{3}}{\sqrt{R^{2} - d^{2}}}}

3

\sqrt{\frac{m g ε_{0} d^{3}}{\sqrt{R^{2} + d^{2}}}}

4

\sqrt{m g ε_{0} d^{3}}

Explanation:

The two charges are shown in the figure.
By considering the equilibrium of any charge
$N \cos θ = \frac{q^{2}}{4 π ε_{0} d^{2}} (1)$
$N \sin θ = m g (2)$
From both the equations
$Tan θ = \frac{m g 4 π ε_{0} d^{2}}{q^{2}} = \frac{\sqrt{R^{2} - \frac{d^{2}}{4}}}{d / 2}$ $q = \sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358055 A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to

1

- \frac{Q}{2}

2

- \frac{Q}{4}

3

+ \frac{Q}{4}

4

\frac{Q}{2}

Explanation:

The central charge is in the equilibrium. If the system has to be in equilibrium then the net force on any of the charge $Q$ should be zero.
$\frac{1}{4 π ε_{0}} \frac{Q Q}{l^{2}} + \frac{1}{4 π ε_{0}} \frac{q Q}{{(\frac{l}{2})}^{2}} = 0$
$\Rightarrow q = \frac{- Q}{4}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358056 Two positively charged spheres of masses, $m_{1} a n d m_{2}$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$ . Charges on the two spheres are $q_{1} a n d q_{2}$ ,respectively. At equilibrium both strings make the same angle $θ$ with the vertical.Then

1

q_{1} m_{1} = q_{2} m_{2}

2

m_{1} = m_{2}

3

m_{1} = m_{2} \sin θ

4

q_{2} m_{1} = q_{1} m_{2}

Explanation:

The forces on the two masses are shown in the figure. Let $θ$ is the angle made by the strings with vertical
supporting img
For equilibrium of $m_{1}$
$T_{1} \cos θ = m_{1} g$
$T_{1} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (1)$
For equilibrium of $m_{2}$
$T_{2} \cos θ = m_{2} g$
$T_{2} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (2)$
From (1) & (2)
$m_{1} = m_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358052 Two small spheres of masses $M_{1} a n d M_{2}$ are suspended by weightless insulating threads of lengths $L_{1} a n d L_{2}$ . The spheres carry charges $Q_{1} a n d Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $θ_{1} a n d θ_{2}$ as shown in figure. Which one of the following conditions is essential, if $θ_{1} = θ_{2} ?$
supporting img

1

M_{1} \neq M_{2} b u t Q_{1} = Q_{2}

2

M_{1} = M_{2}

3

Q_{1} = Q_{2}

4

L_{1} = L_{2}

Explanation:

The three forces acting on each sphere are;
$i) T e n s i o n i i) W e i g h t$
$i i i) E l e c t r o s t a t i c f o r c e o f r e p u l s i o n$
For sphere 1
In equilibrium, from figure,
$T_{1} \cos θ_{1} = M_{1} g; T_{1} \sin θ_{1} = F_{1}$
$∴ \tan θ_{1} = \frac{F_{1}}{M_{1} g}$
For sphere 2
In equilibrium, from figure
$T_{2} \cos θ_{1} = M_{2} g; T_{2} \sin θ_{2} = F_{2}$
$∴ \tan θ_{2} = \frac{F_{2}}{M_{2} g}$
Force of repulsion between two charges are same.
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2} =$ only if
$\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
supporting img

KCET - 2010

PHXII01:ELECTRIC CHARGES AND FIELDS

358053 Assertion :
A charge is lying at the centre of the line joining two similar charges each which are fixed. The system will be in equilibrium if that charge is one fourth of the similar charges.
Reason :
For charge to be in equilibrium, sum of the forces on charge due to rest of the two charges must be zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

If centre - charge is not present, original charges (given to be similar, say: $+ Q$ and $+ Q$ ) will have coulombic repulsion force between them.
supporting img
In order to nullify it and keep system in equilibrium, the centre - charge that to be placed shall be of different sign (here $q$ shall be negative).
$\frac{1}{4 π ϵ_{0}} [\frac{Q Q}{d^{2}}] = \frac{1}{4 π ϵ_{0}} [\frac{Q (- q)}{(d / 2)^{2}}]$
$\Rightarrow | q | = | \frac{Q}{4} | \Rightarrow q = (- Q / 4)$
For the charges to be in equilibrium, the vector sum of the forces due to the rest of the two charges must be zero. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358054 Two small spheres each of mass $m$ and with a charge $q$ lie inside a nonconducting smooth hemispherical bowl of radius $R$ . Find $q$ if the equilibrium separation between the two charges is $d$ .

1

\sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}

2

\sqrt{\frac{m g π ε_{0} d^{3}}{\sqrt{R^{2} - d^{2}}}}

3

\sqrt{\frac{m g ε_{0} d^{3}}{\sqrt{R^{2} + d^{2}}}}

4

\sqrt{m g ε_{0} d^{3}}

Explanation:

The two charges are shown in the figure.
By considering the equilibrium of any charge
$N \cos θ = \frac{q^{2}}{4 π ε_{0} d^{2}} (1)$
$N \sin θ = m g (2)$
From both the equations
$Tan θ = \frac{m g 4 π ε_{0} d^{2}}{q^{2}} = \frac{\sqrt{R^{2} - \frac{d^{2}}{4}}}{d / 2}$ $q = \sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358055 A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to

1

- \frac{Q}{2}

2

- \frac{Q}{4}

3

+ \frac{Q}{4}

4

\frac{Q}{2}

Explanation:

The central charge is in the equilibrium. If the system has to be in equilibrium then the net force on any of the charge $Q$ should be zero.
$\frac{1}{4 π ε_{0}} \frac{Q Q}{l^{2}} + \frac{1}{4 π ε_{0}} \frac{q Q}{{(\frac{l}{2})}^{2}} = 0$
$\Rightarrow q = \frac{- Q}{4}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358056 Two positively charged spheres of masses, $m_{1} a n d m_{2}$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$ . Charges on the two spheres are $q_{1} a n d q_{2}$ ,respectively. At equilibrium both strings make the same angle $θ$ with the vertical.Then

1

q_{1} m_{1} = q_{2} m_{2}

2

m_{1} = m_{2}

3

m_{1} = m_{2} \sin θ

4

q_{2} m_{1} = q_{1} m_{2}

Explanation:

The forces on the two masses are shown in the figure. Let $θ$ is the angle made by the strings with vertical
supporting img
For equilibrium of $m_{1}$
$T_{1} \cos θ = m_{1} g$
$T_{1} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (1)$
For equilibrium of $m_{2}$
$T_{2} \cos θ = m_{2} g$
$T_{2} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (2)$
From (1) & (2)
$m_{1} = m_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358052 Two small spheres of masses $M_{1} a n d M_{2}$ are suspended by weightless insulating threads of lengths $L_{1} a n d L_{2}$ . The spheres carry charges $Q_{1} a n d Q_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $θ_{1} a n d θ_{2}$ as shown in figure. Which one of the following conditions is essential, if $θ_{1} = θ_{2} ?$
supporting img

1

M_{1} \neq M_{2} b u t Q_{1} = Q_{2}

2

M_{1} = M_{2}

3

Q_{1} = Q_{2}

4

L_{1} = L_{2}

Explanation:

The three forces acting on each sphere are;
$i) T e n s i o n i i) W e i g h t$
$i i i) E l e c t r o s t a t i c f o r c e o f r e p u l s i o n$
For sphere 1
In equilibrium, from figure,
$T_{1} \cos θ_{1} = M_{1} g; T_{1} \sin θ_{1} = F_{1}$
$∴ \tan θ_{1} = \frac{F_{1}}{M_{1} g}$
For sphere 2
In equilibrium, from figure
$T_{2} \cos θ_{1} = M_{2} g; T_{2} \sin θ_{2} = F_{2}$
$∴ \tan θ_{2} = \frac{F_{2}}{M_{2} g}$
Force of repulsion between two charges are same.
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2} =$ only if
$\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$
But $F_{1} = F_{2},$ then $M_{1} = M_{2}$
supporting img

KCET - 2010

PHXII01:ELECTRIC CHARGES AND FIELDS

358053 Assertion :
A charge is lying at the centre of the line joining two similar charges each which are fixed. The system will be in equilibrium if that charge is one fourth of the similar charges.
Reason :
For charge to be in equilibrium, sum of the forces on charge due to rest of the two charges must be zero.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

If centre - charge is not present, original charges (given to be similar, say: $+ Q$ and $+ Q$ ) will have coulombic repulsion force between them.
supporting img
In order to nullify it and keep system in equilibrium, the centre - charge that to be placed shall be of different sign (here $q$ shall be negative).
$\frac{1}{4 π ϵ_{0}} [\frac{Q Q}{d^{2}}] = \frac{1}{4 π ϵ_{0}} [\frac{Q (- q)}{(d / 2)^{2}}]$
$\Rightarrow | q | = | \frac{Q}{4} | \Rightarrow q = (- Q / 4)$
For the charges to be in equilibrium, the vector sum of the forces due to the rest of the two charges must be zero. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358054 Two small spheres each of mass $m$ and with a charge $q$ lie inside a nonconducting smooth hemispherical bowl of radius $R$ . Find $q$ if the equilibrium separation between the two charges is $d$ .

1

\sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}

2

\sqrt{\frac{m g π ε_{0} d^{3}}{\sqrt{R^{2} - d^{2}}}}

3

\sqrt{\frac{m g ε_{0} d^{3}}{\sqrt{R^{2} + d^{2}}}}

4

\sqrt{m g ε_{0} d^{3}}

Explanation:

The two charges are shown in the figure.
By considering the equilibrium of any charge
$N \cos θ = \frac{q^{2}}{4 π ε_{0} d^{2}} (1)$
$N \sin θ = m g (2)$
From both the equations
$Tan θ = \frac{m g 4 π ε_{0} d^{2}}{q^{2}} = \frac{\sqrt{R^{2} - \frac{d^{2}}{4}}}{d / 2}$ $q = \sqrt{\frac{2 m g π ε_{0} d^{3}}{\sqrt{R^{2} - \frac{d^{2}}{4}}}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358055 A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to

1

- \frac{Q}{2}

2

- \frac{Q}{4}

3

+ \frac{Q}{4}

4

\frac{Q}{2}

Explanation:

The central charge is in the equilibrium. If the system has to be in equilibrium then the net force on any of the charge $Q$ should be zero.
$\frac{1}{4 π ε_{0}} \frac{Q Q}{l^{2}} + \frac{1}{4 π ε_{0}} \frac{q Q}{{(\frac{l}{2})}^{2}} = 0$
$\Rightarrow q = \frac{- Q}{4}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358056 Two positively charged spheres of masses, $m_{1} a n d m_{2}$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$ . Charges on the two spheres are $q_{1} a n d q_{2}$ ,respectively. At equilibrium both strings make the same angle $θ$ with the vertical.Then

1

q_{1} m_{1} = q_{2} m_{2}

2

m_{1} = m_{2}

3

m_{1} = m_{2} \sin θ

4

q_{2} m_{1} = q_{1} m_{2}

Explanation:

The forces on the two masses are shown in the figure. Let $θ$ is the angle made by the strings with vertical
supporting img
For equilibrium of $m_{1}$
$T_{1} \cos θ = m_{1} g$
$T_{1} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (1)$
For equilibrium of $m_{2}$
$T_{2} \cos θ = m_{2} g$
$T_{2} \sin θ = F$
$\tan θ = \frac{F}{m_{1} g} (2)$
From (1) & (2)
$m_{1} = m_{2}$

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