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Electric Charges and Fields

272132 Which of the following graphs shows the correct variation of force when the distance $r$ between two charges varies?

1

2

3

4

Explanation:

(d) From Coulomb's law $F = \frac{K q_{1} q_{2}}{r^{2}}$ i.e., $F \propto \frac{1}{r^{2}}$ which is correctly shown by graph (d).

NCERT Page-10 / N-7

Electric Charges and Fields

272128 A charge of $4 μ C$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

1

1 μ C

and

3 μ C

2

2 μ C

and

2 μ C

3 0 and

4 μ C

4

1.5 μ C

and

2.5 μ C

Explanation:

(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F = \frac{K q (4 - q)}{r^{2}}$
For $F$ to be maximum, $\frac{d F}{d q} = \frac{K}{r^{2}} [4 - 2 q] = 0 \Rightarrow q = 2$

NCERT Page-12 / N-7

Electric Charges and Fields

272129 Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from midpoint on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is :

1

x = d

2

x = \frac{d}{2}

3

x = \frac{d}{\sqrt{2}}

4

x = \frac{d}{2 \sqrt{2}}

NCERT Page-12/ N-7

Electric Charges and Fields

272133 The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is $(m_{p} = 1836 m_{e})$

1

2 F

2

F

3

1836 F

4

\frac{F}{1836}

Explanation:

(b) Electrostatic force is given by $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$
Here, charge and distance are same. So, force between two protons will be same.

NCERT Page-12 / N-7

Electric Charges and Fields

272132 Which of the following graphs shows the correct variation of force when the distance $r$ between two charges varies?

1

2

3

4

Explanation:

(d) From Coulomb's law $F = \frac{K q_{1} q_{2}}{r^{2}}$ i.e., $F \propto \frac{1}{r^{2}}$ which is correctly shown by graph (d).

NCERT Page-10 / N-7

Electric Charges and Fields

272128 A charge of $4 μ C$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

1

1 μ C

and

3 μ C

2

2 μ C

and

2 μ C

3 0 and

4 μ C

4

1.5 μ C

and

2.5 μ C

Explanation:

(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F = \frac{K q (4 - q)}{r^{2}}$
For $F$ to be maximum, $\frac{d F}{d q} = \frac{K}{r^{2}} [4 - 2 q] = 0 \Rightarrow q = 2$

NCERT Page-12 / N-7

Electric Charges and Fields

272129 Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from midpoint on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is :

1

x = d

2

x = \frac{d}{2}

3

x = \frac{d}{\sqrt{2}}

4

x = \frac{d}{2 \sqrt{2}}

NCERT Page-12/ N-7

Electric Charges and Fields

272133 The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is $(m_{p} = 1836 m_{e})$

1

2 F

2

F

3

1836 F

4

\frac{F}{1836}

Explanation:

(b) Electrostatic force is given by $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$
Here, charge and distance are same. So, force between two protons will be same.

NCERT Page-12 / N-7

Electric Charges and Fields

272132 Which of the following graphs shows the correct variation of force when the distance $r$ between two charges varies?

1

2

3

4

Explanation:

(d) From Coulomb's law $F = \frac{K q_{1} q_{2}}{r^{2}}$ i.e., $F \propto \frac{1}{r^{2}}$ which is correctly shown by graph (d).

NCERT Page-10 / N-7

Electric Charges and Fields

272128 A charge of $4 μ C$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

1

1 μ C

and

3 μ C

2

2 μ C

and

2 μ C

3 0 and

4 μ C

4

1.5 μ C

and

2.5 μ C

Explanation:

(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F = \frac{K q (4 - q)}{r^{2}}$
For $F$ to be maximum, $\frac{d F}{d q} = \frac{K}{r^{2}} [4 - 2 q] = 0 \Rightarrow q = 2$

NCERT Page-12 / N-7

Electric Charges and Fields

272129 Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from midpoint on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is :

1

x = d

2

x = \frac{d}{2}

3

x = \frac{d}{\sqrt{2}}

4

x = \frac{d}{2 \sqrt{2}}

NCERT Page-12/ N-7

Electric Charges and Fields

272133 The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is $(m_{p} = 1836 m_{e})$

1

2 F

2

F

3

1836 F

4

\frac{F}{1836}

Explanation:

(b) Electrostatic force is given by $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$
Here, charge and distance are same. So, force between two protons will be same.

NCERT Page-12 / N-7

Electric Charges and Fields

272132 Which of the following graphs shows the correct variation of force when the distance $r$ between two charges varies?

1

2

3

4

Explanation:

(d) From Coulomb's law $F = \frac{K q_{1} q_{2}}{r^{2}}$ i.e., $F \propto \frac{1}{r^{2}}$ which is correctly shown by graph (d).

NCERT Page-10 / N-7

Electric Charges and Fields

272128 A charge of $4 μ C$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

1

1 μ C

and

3 μ C

2

2 μ C

and

2 μ C

3 0 and

4 μ C

4

1.5 μ C

and

2.5 μ C

Explanation:

(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F = \frac{K q (4 - q)}{r^{2}}$
For $F$ to be maximum, $\frac{d F}{d q} = \frac{K}{r^{2}} [4 - 2 q] = 0 \Rightarrow q = 2$

NCERT Page-12 / N-7

Electric Charges and Fields

272129 Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from midpoint on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is :

1

x = d

2

x = \frac{d}{2}

3

x = \frac{d}{\sqrt{2}}

4

x = \frac{d}{2 \sqrt{2}}

NCERT Page-12/ N-7

Electric Charges and Fields

272133 The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is $(m_{p} = 1836 m_{e})$

1

2 F

2

F

3

1836 F

4

\frac{F}{1836}

Explanation:

(b) Electrostatic force is given by $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$
Here, charge and distance are same. So, force between two protons will be same.

NCERT Page-12 / N-7

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