Question

Two point charge $+q$ and $-q$ are held fixed at $(-d,0)$ and $(d,0)$ respectively of a $x-y$ coordinate system. Then

• A. The electric field $E$ at all points on the $x-$axis has the same direction
• B. Work has to be done in bringing a test charge from $\infty$ to the origin
• C. Electric field at all points on $y-$axis is along $x-$axis
• D. The dipole moment is $2qd$along the $x-$axis

Answer 1

Answer: (C), (D)

The correct options are
C Electric field at all points on $y-$axis is along $x-$axis
D The dipole moment is $2qd$along the $x-$axis

To the left of $+q,$ electric field due to $+q$ will dominate and $\overrightarrow{E}$ will be in -ve $x$ direction.

To the right of $-q,$ electric field due to $-q$ will dominate and $\overrightarrow{E}$ will be in -ve $x$ direction.

And in between $+q$ and $-q,$ when $x\in (-d,0),$ $\overrightarrow{E}$ will in +ve x-direction and when $x\in (0,d)$ $\overrightarrow{E}$ will be in +ve x direction. So, (A) is incorrect.

$\Rightarrow$ Potential at origin $=\frac{k(+9)}{2}+\frac{k(-9)}{2}=0$

So, no work will be done in bringing a test change from infinity to 0.

As we can see that, on y-axis vertical component of $\overrightarrow{E}$ due to $+q$ and $-q$ will be cancelled and net $\overrightarrow{E}$ will be in +ve x direction. Option (C) is correct.

$\Rightarrow$ Dipole moment $=$ ( change ) ( distance between $+q$ and $-q$ )

$=\left(q\right)\left(2d\right)$

$=2qd$ and

Direction will be from $-q$ to $+q.$

Hence, both the options C and D are correct.