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 $4\mu C$ 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 $1\mu C$ and $3\mu C$
2 $2\mu C$ and $2\mu C$
3 0 and $4\mu C$
4 $1.5\mu C$ and $2.5\mu C$
Explanation:
(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F=\frac{Kq\left( 4-q \right)}{{{r}^{2}}}$
For $F$ to be maximum, $\frac{dF}{dq}=\frac{K}{{{r}^{2}}}\left[ 4-2q \right]=0\Rightarrow q=2$
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 $x=\frac{d}{2\sqrt{2}}$
Explanation:
(d)
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 $\left( {{m}_{p}}=1836~{{m}_{e}} \right)$
1 $2~F$
2 $F$
3 $1836~F$
4 $\frac{F}{1836}$
Explanation:
(b) Electrostatic force is given by $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Here, charge and distance are same. So, force between two protons will be same.
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 $4\mu C$ 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 $1\mu C$ and $3\mu C$
2 $2\mu C$ and $2\mu C$
3 0 and $4\mu C$
4 $1.5\mu C$ and $2.5\mu C$
Explanation:
(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F=\frac{Kq\left( 4-q \right)}{{{r}^{2}}}$
For $F$ to be maximum, $\frac{dF}{dq}=\frac{K}{{{r}^{2}}}\left[ 4-2q \right]=0\Rightarrow q=2$
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 $x=\frac{d}{2\sqrt{2}}$
Explanation:
(d)
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 $\left( {{m}_{p}}=1836~{{m}_{e}} \right)$
1 $2~F$
2 $F$
3 $1836~F$
4 $\frac{F}{1836}$
Explanation:
(b) Electrostatic force is given by $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Here, charge and distance are same. So, force between two protons will be same.
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 $4\mu C$ 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 $1\mu C$ and $3\mu C$
2 $2\mu C$ and $2\mu C$
3 0 and $4\mu C$
4 $1.5\mu C$ and $2.5\mu C$
Explanation:
(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F=\frac{Kq\left( 4-q \right)}{{{r}^{2}}}$
For $F$ to be maximum, $\frac{dF}{dq}=\frac{K}{{{r}^{2}}}\left[ 4-2q \right]=0\Rightarrow q=2$
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 $x=\frac{d}{2\sqrt{2}}$
Explanation:
(d)
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 $\left( {{m}_{p}}=1836~{{m}_{e}} \right)$
1 $2~F$
2 $F$
3 $1836~F$
4 $\frac{F}{1836}$
Explanation:
(b) Electrostatic force is given by $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Here, charge and distance are same. So, force between two protons will be same.
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 $4\mu C$ 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 $1\mu C$ and $3\mu C$
2 $2\mu C$ and $2\mu C$
3 0 and $4\mu C$
4 $1.5\mu C$ and $2.5\mu C$
Explanation:
(b) Let distance between the two divided charges be r.
From Coulomb's law, force between two charge,
$F=\frac{Kq\left( 4-q \right)}{{{r}^{2}}}$
For $F$ to be maximum, $\frac{dF}{dq}=\frac{K}{{{r}^{2}}}\left[ 4-2q \right]=0\Rightarrow q=2$
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 $x=\frac{d}{2\sqrt{2}}$
Explanation:
(d)
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 $\left( {{m}_{p}}=1836~{{m}_{e}} \right)$
1 $2~F$
2 $F$
3 $1836~F$
4 $\frac{F}{1836}$
Explanation:
(b) Electrostatic force is given by $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Here, charge and distance are same. So, force between two protons will be same.
