The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density \lambda are kept parallel to each other. In their resulting electric field, point charges q and -q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is (are)

Charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement

Both charges execute simple harmonic motion

Both charges will continue moving in the direction of their displacement

Charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement

Electrostatics PYQ — JEE Advanced 2015 · Shift 1 · Q41 with Solution

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Correct answer: C

Electric field due to an infinite line charge
For an infinite line charge of linear charge density $\lambda$ , the electric field at perpendicular distance $r$ is

E=\frac{\lambda}{2\pi\varepsilon_0 r}

and for a positive line charge, the field is directed away from the line.

Equilibrium position
In both figures, the two identical positive line charges are parallel, so at the midpoint between them, the electric fields due to the two lines are equal and opposite. Hence net electric field is zero there.

Therefore, both $+q$ and $-q$ can be in equilibrium at the midpoint.

Let the two line charges be at $x=-a$ and $x=+a$ . Let the charge be displaced slightly to position $x$ with $|x|<a$ .

Net electric field between the lines
At a point $x$ between the lines:

distance from left line $=a+x$
distance from right line $=a-x$

Field due to left line is toward $+x$ :

E_L=\frac{\lambda}{2\pi\varepsilon_0(a+x)}

Field due to right line is toward $-x$ :

E_R=\frac{\lambda}{2\pi\varepsilon_0(a-x)}

So net field is

E(x)=\frac{\lambda}{2\pi\varepsilon_0(a+x)}-\frac{\lambda}{2\pi\varepsilon_0(a-x)}

E(x)=\frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{a+x}-\frac{1}{a-x}\right)

E(x)=\frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{(a-x)-(a+x)}{a^2-x^2}

E(x)= -\frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{2x}{a^2-x^2}

Thus,

E(x)= -\frac{\lambda x}{\pi\varepsilon_0(a^2-x^2)}

For small displacement $x\ll a$ ,

E(x)\approx -\frac{\lambda}{\pi\varepsilon_0 a^2}x

So the electric field is proportional to $-x$ , i.e. it is a restoring field.

Force on charge $+q$

F_+=qE(x)

For small $x$ ,

F_+ \approx -\frac{q\lambda}{\pi\varepsilon_0 a^2}x

This is of the form

F=-kx

which is the condition for simple harmonic motion.

So, charge $+q$ executes SHM.

Force on charge $-q$

F_- = (-q)E(x)

For small $x$ ,

F_- \approx +\frac{q\lambda}{\pi\varepsilon_0 a^2}x

This force is in the same direction as displacement. Hence the equilibrium is unstable.

So if $-q$ is displaced slightly, it keeps moving further in the direction of displacement.

Checking options

A: Both execute SHM — False
B: Both continue moving in direction of displacement — False
C: $+q$ executes SHM while $-q$ continues moving in direction of displacement — True
D: $-q$ executes SHM while $+q$ continues moving in direction of displacement — False

Final answer
The correct option is:

\boxed{\text{C}}

This matches the stored correct answer.

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View written solutionFree

Correct answer: C

Electric field due to an infinite line charge
For an infinite line charge of linear charge density $\lambda$ , the electric field at perpendicular distance $r$ is

E=\frac{\lambda}{2\pi\varepsilon_0 r}

and for a positive line charge, the field is directed away from the line.

Equilibrium position
In both figures, the two identical positive line charges are parallel, so at the midpoint between them, the electric fields due to the two lines are equal and opposite. Hence net electric field is zero there.

Therefore, both $+q$ and $-q$ can be in equilibrium at the midpoint.

Let the two line charges be at $x=-a$ and $x=+a$ . Let the charge be displaced slightly to position $x$ with $|x|<a$ .

Net electric field between the lines
At a point $x$ between the lines:

distance from left line $=a+x$
distance from right line $=a-x$

Field due to left line is toward $+x$ :

E_L=\frac{\lambda}{2\pi\varepsilon_0(a+x)}

Field due to right line is toward $-x$ :

E_R=\frac{\lambda}{2\pi\varepsilon_0(a-x)}

So net field is

E(x)=\frac{\lambda}{2\pi\varepsilon_0(a+x)}-\frac{\lambda}{2\pi\varepsilon_0(a-x)}

E(x)=\frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{a+x}-\frac{1}{a-x}\right)

E(x)=\frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{(a-x)-(a+x)}{a^2-x^2}

E(x)= -\frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{2x}{a^2-x^2}

Thus,

E(x)= -\frac{\lambda x}{\pi\varepsilon_0(a^2-x^2)}

For small displacement $x\ll a$ ,

E(x)\approx -\frac{\lambda}{\pi\varepsilon_0 a^2}x

So the electric field is proportional to $-x$ , i.e. it is a restoring field.

Force on charge $+q$

F_+=qE(x)

For small $x$ ,

F_+ \approx -\frac{q\lambda}{\pi\varepsilon_0 a^2}x

This is of the form

F=-kx

which is the condition for simple harmonic motion.

So, charge $+q$ executes SHM.

Force on charge $-q$

F_- = (-q)E(x)

For small $x$ ,

F_- \approx +\frac{q\lambda}{\pi\varepsilon_0 a^2}x

This force is in the same direction as displacement. Hence the equilibrium is unstable.

So if $-q$ is displaced slightly, it keeps moving further in the direction of displacement.

Checking options

A: Both execute SHM — False
B: Both continue moving in direction of displacement — False
C: $+q$ executes SHM while $-q$ continues moving in direction of displacement — True
D: $-q$ executes SHM while $+q$ continues moving in direction of displacement — False

Final answer
The correct option is:

\boxed{\text{C}}

This matches the stored correct answer.

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