A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charge q separated by a distance 2 a . For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

Electrostatics PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q52 with Solution

View written solutionFree

Correct answer: B

Field needed for zero force on charge $Q$

For the point charge $Q$ at $P$ to experience no net force, the net electric field at $P$ must be zero:

\vec F = Q\vec E_{\text{net}} = 0 \quad \Rightarrow \quad \vec E_{\text{net}}=0

So the field due to dipole 1 and dipole 2 at $P$ must be equal in magnitude and opposite in direction.

Electric field due to dipole 1 on its axial line

For a dipole made of charges $\pm q$ separated by distance $2a$ , the dipole moment magnitude is

p = q(2a)=2aq

At a point on the axis at distance $r$ from the center, the exact electric field magnitude is

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2}-\frac{q}{(r+a)^2}\right]

Simplify:

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}q\,\frac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2}

(r+a)^2-(r-a)^2 = 4ar

Hence,

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{4aqr}{(r^2-a^2)^2}

Electric field due to dipole 2 on its equatorial line

At a point on the equatorial plane at distance $r$ from the center, the exact field magnitude is

E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{2aq}{(r^2+a^2)^{3/2}}

(Direction is opposite to dipole moment; we only need magnitude here.)

Condition for zero net field

Since the two fields must cancel,

E_{\text{axial}} = E_{\text{equatorial}}

So,

\frac{1}{4\pi\varepsilon_0}\frac{4aqr}{(r^2-a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2aq}{(r^2+a^2)^{3/2}}

Cancel common factors $\dfrac{1}{4\pi\varepsilon_0}$ and $2aq$ :

\frac{2r}{(r^2-a^2)^2} = \frac{1}{(r^2+a^2)^{3/2}}

Rearrange:

2r(r^2+a^2)^{3/2} = (r^2-a^2)^2

Introduce the ratio $x=\dfrac{a}{r}$

Let

x=\frac{a}{r}

Then $a=xr$ . Substitute into the equation:

2r\left(r^2+a^2\right)^{3/2} = (r^2-a^2)^2

2r\left(r^2(1+x^2)\right)^{3/2} = \left(r^2(1-x^2)\right)^2

2r\cdot r^3(1+x^2)^{3/2} = r^4(1-x^2)^2

2(1+x^2)^{3/2} = (1-x^2)^2

We now test the options.

Check the options

Option A: $x=0.5$

\text{LHS} = 2(1+0.25)^{3/2}=2(1.25)^{3/2}\approx 2(1.397)\approx 2.79

\text{RHS}=(1-0.25)^2=(0.75)^2=0.5625

Not equal.

Option B: $x=3$

\text{LHS}=2(1+9)^{3/2}=2(10)^{3/2}=2\cdot 10\sqrt{10}\approx 63.2

\text{RHS}=(1-9)^2=64

These are very close.

Option C: $x=10$

\text{LHS}=2(101)^{3/2}\approx 2\cdot 101\cdot \sqrt{101} \gg 1000

\text{RHS}=(1-100)^2=99^2=9801

Not close.

Option D: $x=20$

Again, values are not close enough compared with option B.

Conclusion

The condition is satisfied approximately for

\frac{a}{r} \approx 3

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

More from Electrostatics

View written solutionFree

Correct answer: B

Field needed for zero force on charge $Q$

For the point charge $Q$ at $P$ to experience no net force, the net electric field at $P$ must be zero:

\vec F = Q\vec E_{\text{net}} = 0 \quad \Rightarrow \quad \vec E_{\text{net}}=0

So the field due to dipole 1 and dipole 2 at $P$ must be equal in magnitude and opposite in direction.

Electric field due to dipole 1 on its axial line

For a dipole made of charges $\pm q$ separated by distance $2a$ , the dipole moment magnitude is

p = q(2a)=2aq

At a point on the axis at distance $r$ from the center, the exact electric field magnitude is

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2}-\frac{q}{(r+a)^2}\right]

Simplify:

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}q\,\frac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2}

(r+a)^2-(r-a)^2 = 4ar

Hence,

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{4aqr}{(r^2-a^2)^2}

Electric field due to dipole 2 on its equatorial line

At a point on the equatorial plane at distance $r$ from the center, the exact field magnitude is

E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{2aq}{(r^2+a^2)^{3/2}}

(Direction is opposite to dipole moment; we only need magnitude here.)

Condition for zero net field

Since the two fields must cancel,

E_{\text{axial}} = E_{\text{equatorial}}

So,

\frac{1}{4\pi\varepsilon_0}\frac{4aqr}{(r^2-a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2aq}{(r^2+a^2)^{3/2}}

Cancel common factors $\dfrac{1}{4\pi\varepsilon_0}$ and $2aq$ :

\frac{2r}{(r^2-a^2)^2} = \frac{1}{(r^2+a^2)^{3/2}}

Rearrange:

2r(r^2+a^2)^{3/2} = (r^2-a^2)^2

Introduce the ratio $x=\dfrac{a}{r}$

Let

x=\frac{a}{r}

Then $a=xr$ . Substitute into the equation:

2r\left(r^2+a^2\right)^{3/2} = (r^2-a^2)^2

2r\left(r^2(1+x^2)\right)^{3/2} = \left(r^2(1-x^2)\right)^2

2r\cdot r^3(1+x^2)^{3/2} = r^4(1-x^2)^2

2(1+x^2)^{3/2} = (1-x^2)^2

We now test the options.

Check the options

Option A: $x=0.5$

\text{LHS} = 2(1+0.25)^{3/2}=2(1.25)^{3/2}\approx 2(1.397)\approx 2.79

\text{RHS}=(1-0.25)^2=(0.75)^2=0.5625

Not equal.

Option B: $x=3$

\text{LHS}=2(1+9)^{3/2}=2(10)^{3/2}=2\cdot 10\sqrt{10}\approx 63.2

\text{RHS}=(1-9)^2=64

These are very close.

Option C: $x=10$

\text{LHS}=2(101)^{3/2}\approx 2\cdot 101\cdot \sqrt{101} \gg 1000

\text{RHS}=(1-100)^2=99^2=9801

Not close.

Option D: $x=20$

Again, values are not close enough compared with option B.

Conclusion

The condition is satisfied approximately for

\frac{a}{r} \approx 3

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

Electrostatics question

Electrostatics question

Option A: $x=0.5$

Option B: $x=3$

Option C: $x=10$

Option D: $x=20$

More from Electrostatics

Electrostatics question

Electrostatics question

Option A: $x=0.5$

Option B: $x=3$

Option C: $x=10$

Option D: $x=20$

More from Electrostatics

Option A: x=0.5x=0.5x=0.5

Option B: x=3x=3x=3

Option C: x=10x=10x=10

Option D: x=20x=20x=20

More from Electrostatics

Option A: x=0.5x=0.5x=0.5

Option B: x=3x=3x=3

Option C: x=10x=10x=10

Option D: x=20x=20x=20

More from Electrostatics

Option A: $x=0.5$

Option B: $x=3$

Option C: $x=10$

Option D: $x=20$

Option A: $x=0.5$

Option B: $x=3$

Option C: $x=10$

Option D: $x=20$