Three infinitely long charged thin sheets are placed as shown in the figure. The magnitude of electric field at point $$P$$ is $$\frac{x\sigma}{\epsilon_o}$$. The value of $$x$$ is _______ (all quantities are measured in SI units).

The electric field ($$E$$) due to an infinitely long thin charged sheet with surface charge density $$\sigma$$ is  $$E = \frac{\sigma}{2\epsilon_0}$$.

Field points away from a positive charge ($$+\sigma$$)

Field points towards a negative charge ($$-\sigma$$)

Sheet 1 (at $$x = -a$$, charge $$-\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_1 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

Sheet 2 (at $$x = a$$, charge $$-2\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_2 = -\frac{2\sigma}{2\epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$$

Sheet 3 (at $$x = 3a$$, charge $$\sigma$$):

Point $$P$$ is to the left of this sheet. Since it is positively charged, the field points left (away from the sheet). $$\vec{E}_3 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

$$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3$$

$$\vec{E}_{net} = \left( -\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} \right) \hat{i}$$

$$\vec{E}_{net} = \left( -\frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} \right) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$$

$$|\vec{E}_{net}| = \frac{2\sigma}{\epsilon_0}$$

$$x = 2$$