272160
Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The point on $x$ axis at which net electric field is zero due to these charges is
1 $8~L$
2 $4~L$
3 $2~L$
4 $L$
Explanation:
(c) Let $P$ is the observation point at a distance $r$ from $-2q$ and at $\left( L+r \right)$ from $+8q$.
Given now, net EFI at $P=0$
$\therefore {{\vec{E}}_{1}}=EFI$ (Electric Field Intensity) at $P$ due to $+8q$
${{\vec{E}}_{2}}=EFI$ (Electric Field Intensity) at $P$ due to $-2q$
$\left| \overrightarrow{{{E}_{1}}} \right|=\left| \overrightarrow{{{E}_{2}}} \right|\therefore \frac{k\left( 8q \right)}{{{(L+r)}^{2}}}=\frac{k\left( 2q \right)}{{{r}^{2}}}\therefore \frac{4}{{{(L+r)}^{2}}}=\frac{1}{{{(r)}^{2}}}$
$4{{r}^{2}}={{(L+r)}^{2}}\Rightarrow 2r=L+r$
$r=L~\therefore P$ is at $x=L+L=2~L$ from origin
NCERT Page-20 / N-16 | CBSE Sample 2021-2022
Electric Charges and Fields
272161
A pendulum bob of mass $m$ carrying a charge $q$ is at rest with its string making an angle $\theta $ with the vertical in a uniform horizontal electric field $E$. The tension in the string is
1 $\frac{mg}{sin\theta }$ and $\frac{qE}{cos\theta }$
2 $\frac{mg}{cos\theta }$ and $\frac{qE}{sin\theta }$
3 $\frac{qE}{mg}$
4 $\frac{mg}{qE}$
Explanation:
(b) In equilibrium,
$Tcos\theta =mg$
$Tsin\theta =qE$
From (1), $T=\frac{mg}{cos\theta }$
From (2), $T=\frac{qE}{sin\theta }$
NCERT Page-20 / N-16
Electric Charges and Fields
272162
A point charge $q=-8.0nC$ is located at the origin. The electric field (in $N{{C}^{-1}}$ ) vector at the point $x=1.2~m,y=-1.6$ $m$, as shown in Fig., is
272163
Figure shows an electric quadrupole, with quadrupole moment $\left( Q=2q{{l}^{2}} \right)$. The electric field at a distance from its centre at the axis of the quadrupole is given by
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Electric Charges and Fields
272160
Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The point on $x$ axis at which net electric field is zero due to these charges is
1 $8~L$
2 $4~L$
3 $2~L$
4 $L$
Explanation:
(c) Let $P$ is the observation point at a distance $r$ from $-2q$ and at $\left( L+r \right)$ from $+8q$.
Given now, net EFI at $P=0$
$\therefore {{\vec{E}}_{1}}=EFI$ (Electric Field Intensity) at $P$ due to $+8q$
${{\vec{E}}_{2}}=EFI$ (Electric Field Intensity) at $P$ due to $-2q$
$\left| \overrightarrow{{{E}_{1}}} \right|=\left| \overrightarrow{{{E}_{2}}} \right|\therefore \frac{k\left( 8q \right)}{{{(L+r)}^{2}}}=\frac{k\left( 2q \right)}{{{r}^{2}}}\therefore \frac{4}{{{(L+r)}^{2}}}=\frac{1}{{{(r)}^{2}}}$
$4{{r}^{2}}={{(L+r)}^{2}}\Rightarrow 2r=L+r$
$r=L~\therefore P$ is at $x=L+L=2~L$ from origin
NCERT Page-20 / N-16 | CBSE Sample 2021-2022
Electric Charges and Fields
272161
A pendulum bob of mass $m$ carrying a charge $q$ is at rest with its string making an angle $\theta $ with the vertical in a uniform horizontal electric field $E$. The tension in the string is
1 $\frac{mg}{sin\theta }$ and $\frac{qE}{cos\theta }$
2 $\frac{mg}{cos\theta }$ and $\frac{qE}{sin\theta }$
3 $\frac{qE}{mg}$
4 $\frac{mg}{qE}$
Explanation:
(b) In equilibrium,
$Tcos\theta =mg$
$Tsin\theta =qE$
From (1), $T=\frac{mg}{cos\theta }$
From (2), $T=\frac{qE}{sin\theta }$
NCERT Page-20 / N-16
Electric Charges and Fields
272162
A point charge $q=-8.0nC$ is located at the origin. The electric field (in $N{{C}^{-1}}$ ) vector at the point $x=1.2~m,y=-1.6$ $m$, as shown in Fig., is
272163
Figure shows an electric quadrupole, with quadrupole moment $\left( Q=2q{{l}^{2}} \right)$. The electric field at a distance from its centre at the axis of the quadrupole is given by
272160
Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The point on $x$ axis at which net electric field is zero due to these charges is
1 $8~L$
2 $4~L$
3 $2~L$
4 $L$
Explanation:
(c) Let $P$ is the observation point at a distance $r$ from $-2q$ and at $\left( L+r \right)$ from $+8q$.
Given now, net EFI at $P=0$
$\therefore {{\vec{E}}_{1}}=EFI$ (Electric Field Intensity) at $P$ due to $+8q$
${{\vec{E}}_{2}}=EFI$ (Electric Field Intensity) at $P$ due to $-2q$
$\left| \overrightarrow{{{E}_{1}}} \right|=\left| \overrightarrow{{{E}_{2}}} \right|\therefore \frac{k\left( 8q \right)}{{{(L+r)}^{2}}}=\frac{k\left( 2q \right)}{{{r}^{2}}}\therefore \frac{4}{{{(L+r)}^{2}}}=\frac{1}{{{(r)}^{2}}}$
$4{{r}^{2}}={{(L+r)}^{2}}\Rightarrow 2r=L+r$
$r=L~\therefore P$ is at $x=L+L=2~L$ from origin
NCERT Page-20 / N-16 | CBSE Sample 2021-2022
Electric Charges and Fields
272161
A pendulum bob of mass $m$ carrying a charge $q$ is at rest with its string making an angle $\theta $ with the vertical in a uniform horizontal electric field $E$. The tension in the string is
1 $\frac{mg}{sin\theta }$ and $\frac{qE}{cos\theta }$
2 $\frac{mg}{cos\theta }$ and $\frac{qE}{sin\theta }$
3 $\frac{qE}{mg}$
4 $\frac{mg}{qE}$
Explanation:
(b) In equilibrium,
$Tcos\theta =mg$
$Tsin\theta =qE$
From (1), $T=\frac{mg}{cos\theta }$
From (2), $T=\frac{qE}{sin\theta }$
NCERT Page-20 / N-16
Electric Charges and Fields
272162
A point charge $q=-8.0nC$ is located at the origin. The electric field (in $N{{C}^{-1}}$ ) vector at the point $x=1.2~m,y=-1.6$ $m$, as shown in Fig., is
272163
Figure shows an electric quadrupole, with quadrupole moment $\left( Q=2q{{l}^{2}} \right)$. The electric field at a distance from its centre at the axis of the quadrupole is given by
272160
Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The point on $x$ axis at which net electric field is zero due to these charges is
1 $8~L$
2 $4~L$
3 $2~L$
4 $L$
Explanation:
(c) Let $P$ is the observation point at a distance $r$ from $-2q$ and at $\left( L+r \right)$ from $+8q$.
Given now, net EFI at $P=0$
$\therefore {{\vec{E}}_{1}}=EFI$ (Electric Field Intensity) at $P$ due to $+8q$
${{\vec{E}}_{2}}=EFI$ (Electric Field Intensity) at $P$ due to $-2q$
$\left| \overrightarrow{{{E}_{1}}} \right|=\left| \overrightarrow{{{E}_{2}}} \right|\therefore \frac{k\left( 8q \right)}{{{(L+r)}^{2}}}=\frac{k\left( 2q \right)}{{{r}^{2}}}\therefore \frac{4}{{{(L+r)}^{2}}}=\frac{1}{{{(r)}^{2}}}$
$4{{r}^{2}}={{(L+r)}^{2}}\Rightarrow 2r=L+r$
$r=L~\therefore P$ is at $x=L+L=2~L$ from origin
NCERT Page-20 / N-16 | CBSE Sample 2021-2022
Electric Charges and Fields
272161
A pendulum bob of mass $m$ carrying a charge $q$ is at rest with its string making an angle $\theta $ with the vertical in a uniform horizontal electric field $E$. The tension in the string is
1 $\frac{mg}{sin\theta }$ and $\frac{qE}{cos\theta }$
2 $\frac{mg}{cos\theta }$ and $\frac{qE}{sin\theta }$
3 $\frac{qE}{mg}$
4 $\frac{mg}{qE}$
Explanation:
(b) In equilibrium,
$Tcos\theta =mg$
$Tsin\theta =qE$
From (1), $T=\frac{mg}{cos\theta }$
From (2), $T=\frac{qE}{sin\theta }$
NCERT Page-20 / N-16
Electric Charges and Fields
272162
A point charge $q=-8.0nC$ is located at the origin. The electric field (in $N{{C}^{-1}}$ ) vector at the point $x=1.2~m,y=-1.6$ $m$, as shown in Fig., is
272163
Figure shows an electric quadrupole, with quadrupole moment $\left( Q=2q{{l}^{2}} \right)$. The electric field at a distance from its centre at the axis of the quadrupole is given by
