Gujarati
1. Electric Charges and Fields
normal

The electric field $E$ is measured at a point $P (0,0, d )$ generated due to various charge distributions and the dependence of $E$ on $d$ is found to be different for different charge distributions. List-$I$ contains different relations between $E$ and $d$. List-$II$ describes different electric charge distributions, along with their locations. Match the functions in List-$I$ with the related charge distributions in List-$II$.

 List-$I$  List-$II$
$E$ is independent of $d$ A point charge $Q$ at the origin
$E \propto \frac{1}{d}$ A small dipole with point charges $Q$ at $(0,0, l)$ and $- Q$ at $(0,0,-l)$. Take $2 l \ll d$.
$E \propto \frac{1}{d^2}$ An infinite line charge coincident with the x-axis, with uniform linear charge density $\lambda$
$E \propto \frac{1}{d^3}$ Two infinite wires carrying uniform linear charge density parallel to the $x$-axis. The one along ( $y=0$, $z =l$ ) has a charge density $+\lambda$ and the one along $( y =0, z =-l)$ has a charge density $-\lambda$. Take $2 l \ll d$
  plane with uniform surface charge density

 

A

$P \rightarrow 5 ; Q \rightarrow 3,4 ; R \rightarrow 1 ; S \rightarrow 2$

B

$P \rightarrow 5 ; Q \rightarrow 3 ; R \rightarrow 1,4 ; S \rightarrow 2$

C

$P \rightarrow 5 ; Q \rightarrow 3 ; R \rightarrow 1,2 ; S \rightarrow 4$

D

$P \rightarrow 4 ; Q \rightarrow 2,3 ; R \rightarrow 1 ; S \rightarrow 5$

(IIT-2018)

Solution

$( P ) \rightarrow(5),( Q ) \rightarrow(3),( R ) \rightarrow(1),(4) ;( S ) \rightarrow(2)$

$(1) E.F.$ due to a point charge at origin.

$E =\frac{ kq }{ d ^2} \Rightarrow E \propto \frac{1}{ d ^2}$

$(2) E.F.$ at any point on axis of dipole

$E =\frac{2 KP }{ d ^3}=\frac{4 KQL }{ d ^3}$

$E \propto \frac{1}{ d ^3}$

$(3) E.F$. due to an infinite long charge

$E =\frac{2 K \lambda}{ d }$

$E \propto \frac{1}{ d }$

$(4) E.F.$ due to two infinite long wires

$\overrightarrow{ E }=\overrightarrow{ E }_1+\overrightarrow{ E }_2$

$\overrightarrow{ E }=\frac{2 K \lambda}{ d – L }-\frac{2 K \lambda}{ d + L }=\frac{2 k \lambda(2 L )}{\left( d ^2- L ^2\right)}=\frac{4 K \lambda L }{ d ^2- L ^2}$

$\text { If } d \gg L \Rightarrow E =\frac{4 K \lambda L }{ d ^2} \Rightarrow E \propto \frac{1}{ d ^2}$

$(5) E.F$. due to infinite plane charge

$E =\frac{\sigma}{\epsilon_0} \quad \text { (independent of } d \text { ) }$

Standard 12
Physics

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