Question

Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3m.

• A. The point P where the resultant field due to both is zero is on AB outside AB
• B. The point P where the resultant field due to both is zero is on AB inside AB
• C. If a positive charge is placed at P and displaced slightly along AB it will execute oscillations
• D. If a negative charge is placed at P and displaced slightly along AB it will execute oscillation

Answer 1

Answer: (A), (D)

(a), (d)

Answer 2

The charges are $+4Q$ at A and $-Q$ at B, with the distance AB = 3m.

First, we find the point P where the electric field is zero. Let P be on the line AB.
If P is between A and B, the electric field due to $+4Q$ is directed towards B, and the electric field due to $-Q$ is directed towards B. Since both fields are in the same direction, the resultant field cannot be zero.

If P is outside the segment AB. Let P be at a distance x from B to the right of B. Then the distance from A is $3+x$.
The electric field at P due to $+4Q$ is $E_A = \frac{k(4Q)}{(3+x)^2}$ directed to the right.
The electric field at P due to $-Q$ is $E_B = \frac{k|-Q|}{x^2} = \frac{kQ}{x^2}$ directed to the left.
For the net field to be zero, the magnitudes must be equal:
$E_A = E_B$
$\frac{k(4Q)}{(3+x)^2} = \frac{kQ}{x^2}$
$\frac{4}{(3+x)^2} = \frac{1}{x^2}$
$4x^2 = (3+x)^2$
Taking the square root of both sides:
$2x = \pm (3+x)$
Case 1: $2x = 3+x \implies x = 3$. This is a valid positive distance.
Case 2: $2x = -(3+x) \implies 2x = -3-x \implies 3x = -3 \implies x = -1$. This is not a valid positive distance.
So, the point P is located at a distance of 3m to the right of B. Since AB = 3m, P is on the line AB and outside the segment AB. This confirms option (a).

If P is to the left of A, let the distance from A be x. Then the distance from B is $3+x$.
The electric field at P due to $+4Q$ is $E_A = \frac{k(4Q)}{x^2}$ directed to the left.
The electric field at P due to $-Q$ is $E_B = \frac{k|-Q|}{(3+x)^2} = \frac{kQ}{(3+x)^2}$ directed to the right.
For the net field to be zero, the magnitudes must be equal:
$E_A = E_B$
$\frac{k(4Q)}{x^2} = \frac{kQ}{(3+x)^2}$
$\frac{4}{x^2} = \frac{1}{(3+x)^2}$
$4(3+x)^2 = x^2$
$2(3+x) = \pm x$
Case 1: $6+2x = x \implies x = -6$. Not a valid positive distance.
Case 2: $6+2x = -x \implies 3x = -6 \implies x = -2$. Not a valid positive distance.
So, the point P is to the right of B.

Thus, the point P where the resultant field is zero is on AB outside AB. Option (a) is correct. Option (b) is incorrect.

Now consider the stability of equilibrium at P. Let the point P be at $x_0 = 6$ (assuming A is at $x=0$ and B is at $x=3$). The electric field at a point x on the line is $E(x) = \frac{k(4Q)}{x^2} - \frac{kQ}{(x-3)^2}$ for $x > 3$.
Consider a small displacement $\delta x$ from P, so $x = x_0 + \delta x = 6 + \delta x$.
$E(6+\delta x) = \frac{k(4Q)}{(6+\delta x)^2} - \frac{kQ}{(3+\delta x)^2}$
For small $\delta x$, we can use Taylor expansion or binomial approximation.
$E(x) = k(4Q)x^{-2} - kQ(x-3)^{-2}$
$E'(x) = k(4Q)(-2)x^{-3} - kQ(-2)(x-3)^{-3}(1) = -8kQx^{-3} + 2kQ(x-3)^{-3}$
At $x=6$, $E'(6) = -8kQ(6)^{-3} + 2kQ(6-3)^{-3} = -8kQ/216 + 2kQ/27 = -kQ/27 + 2kQ/27 = kQ/27$.
Since $Q > 0$, $E'(6) = kQ/27 > 0$.
The force on a test charge $q_0$ is $F(x) = q_0 E(x)$.
For a small displacement $\delta x$ from equilibrium $x_0$, $F(x_0+\delta x) \approx F(x_0) + F'(x_0)\delta x$.
Since $F(x_0) = 0$, $F(x_0+\delta x) \approx F'(x_0)\delta x = q_0 E'(x_0)\delta x$.
The condition for stable equilibrium is that the force is a restoring force, i.e., $F(x_0+\delta x)$ and $\delta x$ have opposite signs.
$q_0 E'(x_0) \delta x < 0$.
$q_0 (kQ/27) \delta x < 0$.

If a positive charge is placed at P, $q_0 > 0$. Then $(kQ/27) \delta x < 0$. Since $kQ/27 > 0$, we need $\delta x < 0$. This means if displaced to the right ($\delta x > 0$), the force is to the right, which is unstable. If displaced to the left ($\delta x < 0$), the force is to the left, which is also unstable. So, for a positive charge, the equilibrium is unstable. Option (c) is incorrect.

If a negative charge is placed at P, $q_0 < 0$. Then $(kQ/27) \delta x > 0$. Since $kQ/27 > 0$, we need $q_0 \delta x > 0$. Since $q_0 < 0$, we need $\delta x < 0$. This means if displaced to the right ($\delta x > 0$), the force is to the left (restoring). If displaced to the left ($\delta x < 0$), the force is to the right (restoring). So, for a negative charge, the equilibrium is stable. If displaced along AB, the force is proportional to the displacement for small displacements, so it will execute oscillations. Option (d) is correct.