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Step 1: Identify the Charges and Their Positions
• A positive charge $+Q$ is fixed at $x = a$.
• A negative charge $-2Q$ is fixed at $x = 2a$.
• We want to find the point(s) on the $x$-axis where the resultant electric field due to these two charges is zero.
Step 2: Express the Electric Fields at a General Point $x$
• The magnitude of the electric field due to a point charge $q$ at distance $r$ is given by $E = \frac{k\,q}{r^2}$, where $k$ is Coulomb's constant.
• Let us consider a general point at coordinate $x$ on the $x$-axis:
The electric field due to $+Q$ (located at $x=a$) at this point is
$E_{+Q} = \frac{kQ}{(x - a)^2}$, directed away from the positive charge.
The electric field due to $-2Q$ (located at $x=2a$) at this point is
$E_{-2Q} = \frac{k(-2Q)}{(x - 2a)^2}$, directed toward the negative charge (since it is negative).
Step 3: Write the Condition for Zero Net Electric Field
For the net electric field to be zero at $x$,
$E_{+Q} + E_{-2Q} = 0.$
Step 4: Equate Magnitudes
We equate the magnitudes of the two fields (ignoring signs for the moment):
$\frac{kQ}{(x - a)^2} = \frac{k(2Q)}{(x - 2a)^2}.$
Cancel $kQ$ (assuming $Q \neq 0$):
$\frac{1}{(x - a)^2} = \frac{2}{(x - 2a)^2}.$
Step 5: Solve for $x$
Cross multiply:
$(x - 2a)^2 = 2 \,(x - a)^2.$
Expand both sides:
Left side: $(x - 2a)^2 = x^2 - 4ax + 4a^2.$
Right side: $2(x - a)^2 = 2\bigl(x^2 - 2ax + a^2\bigr) = 2x^2 - 4ax + 2a^2.$
Equate and simplify:
\[
x^2 - 4ax + 4a^2 = 2x^2 - 4ax + 2a^2
\]
\[
0 = 2x^2 - 4ax + 2a^2 - (x^2 - 4ax + 4a^2) \implies x^2 = 2a^2.
\]
Thus, the solutions are
\[
x = +\sqrt{2}\,a \quad \text{or} \quad x = -\sqrt{2}\,a.
\]
Step 6: Check the Direction of Fields to Confirm the Valid Solution
Upon carefully analyzing the directions of both fields, it can be shown that real cancellation occurs only at
$x = -\sqrt{2}\,a.$
Hence the net electric field is zero only at
$\boxed{x = -\sqrt{2}\,a}.$
