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Electric Field

A few electri...

Question

A few electric field lines for a system of two charges $Q_{1}$ and $Q_{2}$ fixed at two different points on the $x -$ axis are shown in the figure. These lines suggest that

A

| Q_{1} | > | Q_{2} |

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B

| Q_{1} | < | Q_{2} |

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C

at a finite distance to the left of

Q_{1}

the electric field is zero

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D

at a finite distance to the right of

Q_{2}

the electric field is zero

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Solution

The correct options are
A $| Q_{1} | > | Q_{2} |$
B at a finite distance to the right of $Q_{2}$ the electric field is zero
As electric field lines are originating from $Q_{1}$ and terminating at $Q_{2}$
$\Rightarrow Q_{1}$ is +ve and $Q_{2}$ is (-)ve.
Density of field lines $α$ strength of electric field $α$ $| Q |$
So, $| Q_{1} | > | Q_{2} |$ (A)

$\Rightarrow E_{1} =$ Electric field at P due to $Q_{1}$
$E_{2} =$ Electric field at P due to $Q_{2}$
For $E_{p} = 0, E_{1} - E_{2} = 0$
$\Rightarrow E_{1} - E_{2} = 0$
$\Rightarrow E_{1} = E_{2}$
$\Rightarrow \frac{k Q_{1}}{{(r + d)}^{2}} = \frac{k Q_{2}}{r^{2}}$
$\Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{{(r + d)}^{2}}{r^{2}}$
As $| Q_{1} | > | Q_{2} |,$ so, such $^{'} r^{'}$ exists.
Hence, both option (A) and (D) are correct.

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