In the (x,y,z) coordinate system, three point-charges Q, Q, and αQ are located in free space at (−1, 0, 0), (1, 0, 0) and (0,−1,0), respectively. The value of α for the electric field to be zero at (0, 0.5, 0) is ____________ (rounded off to 1 decimal place).

The correct answer is -1.6.

To find the value of $\alpha$ such that the total electric field at the point $P(0,0.5,0)$ is zero, we calculate the electric field contribution from each of the three point-charges.

Let the electrostatic constant be $k=\frac{1}{4\pi {\epsilon }_0}$.

1. Electric field due to the charge $Q$ at $r_1=(-1,0,0)$:
The position vector from the charge to the point $P$ is:
${\overrightarrow{R}}_1=(0-(-1\left)\right)\hat{i}+(0.5-0)\hat{j}=\hat{i}+0.5\hat{j}$
The distance is:
$R_1=\sqrt{1^2+{0.5}^2}=\sqrt{1.25}$
The electric field vector ${\overrightarrow{E}}_1$ is:
${\overrightarrow{E}}_1=\frac{kQ}{{R_1}^3}{\overrightarrow{R}}_1=\frac{kQ}{{1.25}^{1.5}}(\hat{i}+0.5\hat{j})$

2. Electric field due to the charge $Q$ at $r_2=(1,0,0)$:
The position vector from the charge to the point $P$ is:
${\overrightarrow{R}}_2=(0-1)\hat{i}+(0.5-0)\hat{j}=-\hat{i}+0.5\hat{j}$
The distance is:
$R_2=\sqrt{{(-1)}^2+{0.5}^2}=\sqrt{1.25}$
The electric field vector ${\overrightarrow{E}}_2$ is:
${\overrightarrow{E}}_2=\frac{kQ}{{R_2}^3}{\overrightarrow{R}}_2=\frac{kQ}{{1.25}^{1.5}}(-\hat{i}+0.5\hat{j})$

3. Combined electric field of the first two charges:
Adding the two fields ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$, the horizontal components cancel out, and we get:
${\overrightarrow{E}}_{12}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\frac{kQ}{{1.25}^{1.5}}(0.5+0.5)\hat{j}=\frac{kQ}{{1.25}^{1.5}}\hat{j}$

4. Electric field due to the charge $\alpha Q$ at $r_3=(0,-1,0)$:
The position vector from this charge to the point $P$ is:
${\overrightarrow{R}}_3=(0-0)\hat{i}+(0.5-(-1\left)\right)\hat{j}=1.5\hat{j}$
The distance is:
$R_3=1.5$
The electric field vector ${\overrightarrow{E}}_3$ is:
${\overrightarrow{E}}_3=\frac{k\left(\alpha Q\right)}{{R_3}^3}{\overrightarrow{R}}_3=\frac{k\alpha Q}{{1.5}^2}\hat{j}=\frac{k\alpha Q}{2.25}$

5. Equating the total electric field to zero:
${\overrightarrow{E}}_{\mathrm{total}}={\overrightarrow{E}}_{12}+{\overrightarrow{E}}_3=0$
$(\frac{kQ}{{1.25}^{1.5}}+\frac{k\alpha Q}{2.25})\hat{j}=0$
Dividing both sides by $kQ$:
$\frac{1}{{1.25}^{1.5}}+\frac{\alpha }{2.25}=0$
$\alpha =-\frac{2.25}{{1.25}^{1.5}}$
Evaluating the value of ${1.25}^{1.5}$:
${1.25}^{1.5}\approx 1.3975$
Now substituting this back to find $\alpha$:
$\alpha \approx -\frac{2.25}{1.3975}\approx -1.61$

Rounding off to 1 decimal place, we get:
$\alpha =-1.6$