Two charges, each equals to q, are kept at x=-a and x=a on the x -axis. A particle of mass m and charge q_0 = q \over 2 is placed at the origin. If charge q_0 is given a small displacement ( y \lt \lt a ) along the y -axis, the net force acting on the particle is proportional to

Electrostatics PYQ — JEE Main 2013 · Shift 0 · Q65 with Solution

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Correct answer: A

Set up the configuration

Two fixed charges $q$ are at

\quad \text{and} \quad (a,0).$$ A particle of charge $$q_0=\frac q2$$ is initially at the origin and then displaced slightly to $$(0,y), \qquad y\ll a.$$ We need the net force on $q_0$ and determine how it depends on $y$. --- 2. **Distance from displaced charge to each fixed charge** From $(0,y)$ to either fixed charge, the distance is $$r=\sqrt{a^2+y^2}.$$ So the magnitude of force due to one fixed charge on $q_0$ is $$F_1=\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{r^2} =\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{a^2+y^2}.$$ --- 3. **Resolve force into components** Because the arrangement is symmetric, the $x$-components of the two forces cancel. The $y$-component from one charge is $$F_{1y}=F_1\frac{y}{r} =\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{a^2+y^2}\cdot \frac{y}{\sqrt{a^2+y^2}}.

Thus, $F_{1y}=\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0\,y}{(a^2+y^2)^{3/2}}.$

Since there are two identical charges, total force along $y$ is $F_y=2\cdot \frac{1}{4\pi\varepsilon_0}\frac{q\,q_0\,y}{(a^2+y^2)^{3/2}}.$

So, $F_y=\frac{1}{4\pi\varepsilon_0}\frac{2q q_0 y}{(a^2+y^2)^{3/2}}.$

Use small displacement condition $y\ll a$

For $y\ll a$ , $(a^2+y^2)^{3/2}\approx a^3.$

Hence, $F_y\approx \frac{1}{4\pi\varepsilon_0}\frac{2q q_0}{a^3}\, y.$

Since $q_0=\dfrac q2$ , $F_y\approx \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^3}\, y.$

Therefore, $F\propto y.$

Direction check

If all charges are positive, the displaced charge is repelled upward by both fixed charges, so the force is in the same direction as $+y$ . Thus it is proportional to $y$ , not $-y$ .

Evaluate options

A: $y$ ✅ Correct
B: $-y$ ❌ Wrong sign
C: $1/y$ ❌ Not obtained
D: $-1/y$ ❌ Not obtained

Hence the correct answer is A.

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View written solutionFree

Correct answer: A

Set up the configuration

Two fixed charges $q$ are at

\quad \text{and} \quad (a,0).$$ A particle of charge $$q_0=\frac q2$$ is initially at the origin and then displaced slightly to $$(0,y), \qquad y\ll a.$$ We need the net force on $q_0$ and determine how it depends on $y$. --- 2. **Distance from displaced charge to each fixed charge** From $(0,y)$ to either fixed charge, the distance is $$r=\sqrt{a^2+y^2}.$$ So the magnitude of force due to one fixed charge on $q_0$ is $$F_1=\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{r^2} =\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{a^2+y^2}.$$ --- 3. **Resolve force into components** Because the arrangement is symmetric, the $x$-components of the two forces cancel. The $y$-component from one charge is $$F_{1y}=F_1\frac{y}{r} =\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0}{a^2+y^2}\cdot \frac{y}{\sqrt{a^2+y^2}}.

Thus, $F_{1y}=\frac{1}{4\pi\varepsilon_0}\frac{q\,q_0\,y}{(a^2+y^2)^{3/2}}.$

Since there are two identical charges, total force along $y$ is $F_y=2\cdot \frac{1}{4\pi\varepsilon_0}\frac{q\,q_0\,y}{(a^2+y^2)^{3/2}}.$

So, $F_y=\frac{1}{4\pi\varepsilon_0}\frac{2q q_0 y}{(a^2+y^2)^{3/2}}.$

Use small displacement condition $y\ll a$

For $y\ll a$ , $(a^2+y^2)^{3/2}\approx a^3.$

Hence, $F_y\approx \frac{1}{4\pi\varepsilon_0}\frac{2q q_0}{a^3}\, y.$

Since $q_0=\dfrac q2$ , $F_y\approx \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^3}\, y.$

Therefore, $F\propto y.$

Direction check

If all charges are positive, the displaced charge is repelled upward by both fixed charges, so the force is in the same direction as $+y$ . Thus it is proportional to $y$ , not $-y$ .

Evaluate options

A: $y$ ✅ Correct
B: $-y$ ❌ Wrong sign
C: $1/y$ ❌ Not obtained
D: $-1/y$ ❌ Not obtained

Hence the correct answer is A.

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