Question

If a dipole of dipole moment $p\hat{i}$ is placed at point (0, y) and a point charge at the origin of a coordinate system, net electric field at point (x, x + y) vanishes. If x and y both are positive, the coordinate y is equal to

• A. x
• B. 2x
• C. 2.5x
• D. 3x

Answer 1

Answer: (B)

2x

Answer 2

The problem requires us to find the relationship between x and y such that the net electric field at point P(x, x+y) is zero, given a point charge at the origin and a dipole at (0, y).

1. Electric Field due to the Point Charge (Q) at the Origin (0, 0): Let the point charge be Q. The position vector of point P(x, x+y) from the origin is $\vec{r}_Q = x\hat{i} + (x+y)\hat{j}$. The magnitude of this vector is $r_Q = \sqrt{x^2 + (x+y)^2}$. The electric field due to the point charge Q at P is: $\vec{E}_Q = \frac{1}{4\pi\epsilon_0} \frac{Q}{r_Q^3} \vec{r}_Q = K \frac{Q}{(x^2 + (x+y)^2)^{3/2}} (x\hat{i} + (x+y)\hat{j})$ where $K = \frac{1}{4\pi\epsilon_0}$.

2. Electric Field due to the Dipole ($\vec{p} = p\hat{i}$) at (0, y): The dipole is located at (0, y). The position vector of point P(x, x+y) relative to the dipole's center is: $\vec{r}_p = (x - 0)\hat{i} + ((x+y) - y)\hat{j} = x\hat{i} + x\hat{j}$. The magnitude of this vector is $r_p = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2}$. The dipole moment is $\vec{p} = p\hat{i}$.

The general formula for the electric field due to a dipole $\vec{p}$ at a position vector $\vec{r}$ from its center is: $\vec{E}_p = \frac{1}{4\pi\epsilon_0} \frac{1}{r^5} [3(\vec{p} \cdot \vec{r})\vec{r} - \vec{p}r^2]$ Substituting the values: $\vec{p} \cdot \vec{r}_p = (p\hat{i}) \cdot (x\hat{i} + x\hat{j}) = px$. $r_p^2 = (x\sqrt{2})^2 = 2x^2$.

$\vec{E}_p = K \frac{1}{(x\sqrt{2})^5} [3(px)(x\hat{i} + x\hat{j}) - p\hat{i}(2x^2)]$ $\vec{E}_p = K \frac{1}{4\sqrt{2}x^5} [3px^2(\hat{i} + \hat{j}) - 2px^2\hat{i}]$ $\vec{E}_p = K \frac{px^2}{4\sqrt{2}x^5} [(3-2)\hat{i} + 3\hat{j}]$ $\vec{E}_p = K \frac{p}{4\sqrt{2}x^3} (\hat{i} + 3\hat{j})$.

3. Net Electric Field is Zero: The net electric field at P is $\vec{E}_{net} = \vec{E}_Q + \vec{E}_p = \vec{0}$. This implies $\vec{E}_Q = -\vec{E}_p$.

So, $K \frac{Q}{(x^2 + (x+y)^2)^{3/2}} (x\hat{i} + (x+y)\hat{j}) = -K \frac{p}{4\sqrt{2}x^3} (\hat{i} + 3\hat{j})$.

Let $A = \frac{Q}{(x^2 + (x+y)^2)^{3/2}}$ and $B = -\frac{p}{4\sqrt{2}x^3}$. The equation becomes $A(x\hat{i} + (x+y)\hat{j}) = B(\hat{i} + 3\hat{j})$.

Equating the components: x-component: $Ax = B$ (Equation 1) y-component: $A(x+y) = 3B$ (Equation 2)

Substitute the value of B from Equation 1 into Equation 2: $A(x+y) = 3(Ax)$ Since x and y are positive, A cannot be zero (as p is non-zero). Therefore, we can divide by A: $x+y = 3x$ $y = 3x - x$ $y = 2x$.

The coordinate y is equal to 2x.