### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$ axis at which the net electric field due to these two point charges is zero is
A
${L \over 4}$
B
$2$ $L$
C
$4$ $L$
D
$8$ $L$

## Explanation

${{ - K2q} \over {{{\left( {x - L} \right)}^2}}} + {{K8q} \over {{x^2}}} = 0 \Rightarrow {1 \over {{{\left( {x - L} \right)}^2}}} = {4 \over {{x^2}}}$

or, ${1 \over {x - L}} = {2 \over x} \Rightarrow x = 2x - 2L$

or, $x=2L$
2

### AIEEE 2005

A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $'C'$ then the resultant capacitance is
A
$\left( {n + 1} \right)C$
B
$\left( {n - 1} \right)C$
C
$nC$
D
$C$

## Explanation

As $n$ plates are joined, it means $(n-1)$ capacitor joined in parallel.

$\therefore$ resultant capacitance $=(n-1)C$
3

### AIEEE 2004

A charged oil drop is suspended in a uniform field of $3 \times {10^4}$ $v/m$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $= 9.9 \times {10^{ - 15}}\,\,kg$ and $g = 10\,m/{s^2}$)
A
$1.6 \times {10^{ - 18}}\,C$
B
$3.2 \times {10^{ - 18}}\,C$
C
$3.3 \times {10^{ - 18}}\,C$
D
$4.8 \times {10^{ - 18}}\,C$

## Explanation

At equilibrium, electric force on drop balances weight of drop.

$qE = mg \Rightarrow q$

$= {{mg} \over E} = {{9.9 \times {{10}^{ - 15}} \times 10} \over {3 \times {{10}^4}}}$

$= 3.3 \times {10^{ - 18}}C$
4

### AIEEE 2004

Four charges equal to -$Q$ are placed at the four corners of a square and a charge $q$ is at its center. If the system is in equilibrium the value of $q$ is
A
$- {Q \over 2}\left( {1 + 2\sqrt 2 } \right)$
B
${Q \over 4}\left( {1 + 2\sqrt 2 } \right)$
C
$- {Q \over 4}\left( {1 + 2\sqrt 2 } \right)$
D
${Q \over 2}\left( {1 + 2\sqrt 2 } \right)$

## Explanation

Net field at A should be zero

$\sqrt 2 \,{E_1} + {E_2} = {E_3}$

$\therefore$ ${{kQ \times \sqrt 2 } \over {{a^2}}} + {{kQ} \over {\left( {\sqrt 2 a} \right)}} = {{kq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}}$

$\Rightarrow {{Q\sqrt 2 } \over 1} + {Q \over 2} = 2q$

$\Rightarrow q = {Q \over 4}\left( {2\sqrt 2 + 1} \right).$