272143
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
(d) ${{Q}_{1}}+{{Q}_{2}}=Q$
... (i) and $F=k\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$
From (i) and (ii) $F=\frac{k{{Q}_{1}}\left( Q-{{Q}_{1}} \right)}{{{r}^{2}}}$
For $F$ to be maximum $\frac{dF}{d{{Q}_{1}}}=0\Rightarrow {{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}$
NCERT Page-12 / N-7
Electric Charges and Fields
272144
Two particles $A$ and $B$ having equal charges are placed at a distance $d$ apart. A third charged particle placed on the perpendicular bisection of $AB$ at distance $x$. The third particle experiences maximum force when -
1 $x=\frac{d}{\sqrt{2}}$
2 $x=\frac{d}{2}$
3 $x=\frac{d}{2\sqrt{2}}$
4 $x=\frac{d}{3\sqrt{2}}$
Explanation:
(c)
NCERT Page-12 / N-7
Electric Charges and Fields
272145
Two point charges placed in a medium of dielectric constant 5 are at a distance $r$ between them, experience an electrostatic force ' $F$ '. The electrostatic force between them in vacuum at the same distance $r$ will be-
1 $5~F$
2 $F$
3 $F/2$
4 $F/5$
Explanation:
(a) $\frac{{{Q}_{1}}~{{Q}_{2}}}{r}K=5~~F=\frac{1}{4\pi {{\varepsilon }_{0}}k}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$ ${{Q}_{1}}~{{Q}_{2}}$ force in the charges in the air is ${F}'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}=KF=5~F$
NCERT Page-12 / N-7 I CBSE Sample 2021-2022
Electric Charges and Fields
272146
In fig., two equal positive point charges ${{q}_{1}}={{q}_{2}}=2.0\mu C$ interact with a third point charge $Q=4.0\mu C$. The magnitude, as well as direction, of the net force on $Q$ is
272143
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
(d) ${{Q}_{1}}+{{Q}_{2}}=Q$
... (i) and $F=k\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$
From (i) and (ii) $F=\frac{k{{Q}_{1}}\left( Q-{{Q}_{1}} \right)}{{{r}^{2}}}$
For $F$ to be maximum $\frac{dF}{d{{Q}_{1}}}=0\Rightarrow {{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}$
NCERT Page-12 / N-7
Electric Charges and Fields
272144
Two particles $A$ and $B$ having equal charges are placed at a distance $d$ apart. A third charged particle placed on the perpendicular bisection of $AB$ at distance $x$. The third particle experiences maximum force when -
1 $x=\frac{d}{\sqrt{2}}$
2 $x=\frac{d}{2}$
3 $x=\frac{d}{2\sqrt{2}}$
4 $x=\frac{d}{3\sqrt{2}}$
Explanation:
(c)
NCERT Page-12 / N-7
Electric Charges and Fields
272145
Two point charges placed in a medium of dielectric constant 5 are at a distance $r$ between them, experience an electrostatic force ' $F$ '. The electrostatic force between them in vacuum at the same distance $r$ will be-
1 $5~F$
2 $F$
3 $F/2$
4 $F/5$
Explanation:
(a) $\frac{{{Q}_{1}}~{{Q}_{2}}}{r}K=5~~F=\frac{1}{4\pi {{\varepsilon }_{0}}k}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$ ${{Q}_{1}}~{{Q}_{2}}$ force in the charges in the air is ${F}'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}=KF=5~F$
NCERT Page-12 / N-7 I CBSE Sample 2021-2022
Electric Charges and Fields
272146
In fig., two equal positive point charges ${{q}_{1}}={{q}_{2}}=2.0\mu C$ interact with a third point charge $Q=4.0\mu C$. The magnitude, as well as direction, of the net force on $Q$ is
272143
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
(d) ${{Q}_{1}}+{{Q}_{2}}=Q$
... (i) and $F=k\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$
From (i) and (ii) $F=\frac{k{{Q}_{1}}\left( Q-{{Q}_{1}} \right)}{{{r}^{2}}}$
For $F$ to be maximum $\frac{dF}{d{{Q}_{1}}}=0\Rightarrow {{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}$
NCERT Page-12 / N-7
Electric Charges and Fields
272144
Two particles $A$ and $B$ having equal charges are placed at a distance $d$ apart. A third charged particle placed on the perpendicular bisection of $AB$ at distance $x$. The third particle experiences maximum force when -
1 $x=\frac{d}{\sqrt{2}}$
2 $x=\frac{d}{2}$
3 $x=\frac{d}{2\sqrt{2}}$
4 $x=\frac{d}{3\sqrt{2}}$
Explanation:
(c)
NCERT Page-12 / N-7
Electric Charges and Fields
272145
Two point charges placed in a medium of dielectric constant 5 are at a distance $r$ between them, experience an electrostatic force ' $F$ '. The electrostatic force between them in vacuum at the same distance $r$ will be-
1 $5~F$
2 $F$
3 $F/2$
4 $F/5$
Explanation:
(a) $\frac{{{Q}_{1}}~{{Q}_{2}}}{r}K=5~~F=\frac{1}{4\pi {{\varepsilon }_{0}}k}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$ ${{Q}_{1}}~{{Q}_{2}}$ force in the charges in the air is ${F}'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}=KF=5~F$
NCERT Page-12 / N-7 I CBSE Sample 2021-2022
Electric Charges and Fields
272146
In fig., two equal positive point charges ${{q}_{1}}={{q}_{2}}=2.0\mu C$ interact with a third point charge $Q=4.0\mu C$. The magnitude, as well as direction, of the net force on $Q$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electric Charges and Fields
272143
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
(d) ${{Q}_{1}}+{{Q}_{2}}=Q$
... (i) and $F=k\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$
From (i) and (ii) $F=\frac{k{{Q}_{1}}\left( Q-{{Q}_{1}} \right)}{{{r}^{2}}}$
For $F$ to be maximum $\frac{dF}{d{{Q}_{1}}}=0\Rightarrow {{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}$
NCERT Page-12 / N-7
Electric Charges and Fields
272144
Two particles $A$ and $B$ having equal charges are placed at a distance $d$ apart. A third charged particle placed on the perpendicular bisection of $AB$ at distance $x$. The third particle experiences maximum force when -
1 $x=\frac{d}{\sqrt{2}}$
2 $x=\frac{d}{2}$
3 $x=\frac{d}{2\sqrt{2}}$
4 $x=\frac{d}{3\sqrt{2}}$
Explanation:
(c)
NCERT Page-12 / N-7
Electric Charges and Fields
272145
Two point charges placed in a medium of dielectric constant 5 are at a distance $r$ between them, experience an electrostatic force ' $F$ '. The electrostatic force between them in vacuum at the same distance $r$ will be-
1 $5~F$
2 $F$
3 $F/2$
4 $F/5$
Explanation:
(a) $\frac{{{Q}_{1}}~{{Q}_{2}}}{r}K=5~~F=\frac{1}{4\pi {{\varepsilon }_{0}}k}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$ ${{Q}_{1}}~{{Q}_{2}}$ force in the charges in the air is ${F}'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}=KF=5~F$
NCERT Page-12 / N-7 I CBSE Sample 2021-2022
Electric Charges and Fields
272146
In fig., two equal positive point charges ${{q}_{1}}={{q}_{2}}=2.0\mu C$ interact with a third point charge $Q=4.0\mu C$. The magnitude, as well as direction, of the net force on $Q$ is
