Two identical positive charges Q each are fixed at a distance of '2a' apart from each other. Another point charge q_0 with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q_0 executes SHM . The time period of oscillation of charge q_0 will be :

\sqrt\fracq_0 Q4 \pi^3 \varepsilon_0 m a^3

\sqrt\frac2 \pi^2 \varepsilon_0 m a^3q_0 Q

\sqrt\frac8 \pi^3 \varepsilon_0 m a^3q_0 Q

Electrostatics PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q56 with Solution

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Correct answer: A

Set up the system

Two fixed charges $+Q$ are placed at $x=-a$ and $x=+a$ , since they are $2a$ apart. A charge $q_0$ of mass $m$ is placed at the midpoint $x=0$ .

For SHM along the line joining the charges, the displaced charge must experience a restoring force. So we consider a small displacement $x$ from the midpoint along the $x$ -axis.

Force on $q_0$ after displacement

Suppose $q_0$ is displaced slightly to the right by distance $x$ $(x \ll a)$ .

Then its distances from the two fixed charges are:

from left charge at $-a$ : $a+x$
from right charge at $+a$ : $a-x$

Electric force due to left charge: $F_L = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{(a+x)^2}$ acting toward the right if $q_0>0$ .

Electric force due to right charge: $F_R = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{(a-x)^2}$ acting toward the left if $q_0>0$ .

Hence net force on $q_0$ is $F = \frac{1}{4\pi\varepsilon_0}Qq_0\left[\frac{1}{(a+x)^2}-\frac{1}{(a-x)^2}\right].$

Simplify for small displacement

Using

= \frac{(a-x)^2-(a+x)^2}{(a^2-x^2)^2} = \frac{-4ax}{(a^2-x^2)^2},$$ we get $$F = -\frac{1}{4\pi\varepsilon_0}\frac{4aQq_0x}{(a^2-x^2)^2}.$$ For small $x$, $a^2-x^2 \approx a^2$, so $$(a^2-x^2)^2 \approx a^4.$$ Thus $$F \approx -\frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}x.$$ So the motion is of the form $$F=-kx,$$ with effective spring constant $$k=\frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}.$$ 4. **Condition for SHM** For this force to be restoring, the coefficient of $x$ in $F=-kx$ must be positive. This requires $q_0>0$ in the above expression as written. Hence the given formula assumes a restoring setup and uses the magnitude accordingly. 5. **Angular frequency and time period** For SHM, $$\omega=\sqrt{\frac{k}{m}}= \sqrt{\frac{1}{m}\cdot \frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}}.$$ Now, $$\omega^2=\frac{Qq_0}{\pi\varepsilon_0 m a^3}.$$ Therefore, $$T=\frac{2\pi}{\omega} =2\pi\sqrt{\frac{\pi\varepsilon_0 m a^3}{Qq_0}}.$$ Squaring, $$T^2=4\pi^2\cdot \frac{\pi\varepsilon_0 m a^3}{Qq_0} =\frac{4\pi^3\varepsilon_0 m a^3}{Qq_0}.$$ Hence, $$T=\sqrt{\frac{4\pi^3\varepsilon_0 m a^3}{q_0Q}}.$$ 6. **Match with options** This matches **Option A**. > Note: Physically, for two fixed positive charges, a charge displaced along the line joining them experiences a restoring force only if $q_0$ is negative. The formula for time period uses the magnitude $|q_0|$, but among the given options, the intended answer is still Option A.

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View written solutionFree

Correct answer: A

Set up the system

Two fixed charges $+Q$ are placed at $x=-a$ and $x=+a$ , since they are $2a$ apart. A charge $q_0$ of mass $m$ is placed at the midpoint $x=0$ .

For SHM along the line joining the charges, the displaced charge must experience a restoring force. So we consider a small displacement $x$ from the midpoint along the $x$ -axis.

Force on $q_0$ after displacement

Suppose $q_0$ is displaced slightly to the right by distance $x$ $(x \ll a)$ .

Then its distances from the two fixed charges are:

from left charge at $-a$ : $a+x$
from right charge at $+a$ : $a-x$

Electric force due to left charge: $F_L = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{(a+x)^2}$ acting toward the right if $q_0>0$ .

Electric force due to right charge: $F_R = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{(a-x)^2}$ acting toward the left if $q_0>0$ .

Hence net force on $q_0$ is $F = \frac{1}{4\pi\varepsilon_0}Qq_0\left[\frac{1}{(a+x)^2}-\frac{1}{(a-x)^2}\right].$

Simplify for small displacement

Using

= \frac{(a-x)^2-(a+x)^2}{(a^2-x^2)^2} = \frac{-4ax}{(a^2-x^2)^2},$$ we get $$F = -\frac{1}{4\pi\varepsilon_0}\frac{4aQq_0x}{(a^2-x^2)^2}.$$ For small $x$, $a^2-x^2 \approx a^2$, so $$(a^2-x^2)^2 \approx a^4.$$ Thus $$F \approx -\frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}x.$$ So the motion is of the form $$F=-kx,$$ with effective spring constant $$k=\frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}.$$ 4. **Condition for SHM** For this force to be restoring, the coefficient of $x$ in $F=-kx$ must be positive. This requires $q_0>0$ in the above expression as written. Hence the given formula assumes a restoring setup and uses the magnitude accordingly. 5. **Angular frequency and time period** For SHM, $$\omega=\sqrt{\frac{k}{m}}= \sqrt{\frac{1}{m}\cdot \frac{1}{4\pi\varepsilon_0}\frac{4Qq_0}{a^3}}.$$ Now, $$\omega^2=\frac{Qq_0}{\pi\varepsilon_0 m a^3}.$$ Therefore, $$T=\frac{2\pi}{\omega} =2\pi\sqrt{\frac{\pi\varepsilon_0 m a^3}{Qq_0}}.$$ Squaring, $$T^2=4\pi^2\cdot \frac{\pi\varepsilon_0 m a^3}{Qq_0} =\frac{4\pi^3\varepsilon_0 m a^3}{Qq_0}.$$ Hence, $$T=\sqrt{\frac{4\pi^3\varepsilon_0 m a^3}{q_0Q}}.$$ 6. **Match with options** This matches **Option A**. > Note: Physically, for two fixed positive charges, a charge displaced along the line joining them experiences a restoring force only if $q_0$ is negative. The formula for time period uses the magnitude $|q_0|$, but among the given options, the intended answer is still Option A.

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