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Step-by-Step Solution
Step 1: Understand the Problem
We have three charges placed on the x-axis at positions:
A charge +Q at x = 0
A charge q at x = d/2
A charge +Q at x = d
The net electrostatic force on the charge +Q placed at x = 0 is zero. We aim to find the value of q that satisfies this condition.
Step 2: Identify the Forces Acting on +Q at x = 0
There are two forces acting on this charge:
The force due to the charge q at x = d/2
The force due to the charge +Q at x = d
Step 3: Write the Magnitudes of the Forces Using Coulomb's Law
Coulomb’s law states that the magnitude of force between two point charges $Q_1$ and $Q_2$, separated by a distance $r$, is given by:
$ F = k \frac{Q_1 \, Q_2}{r^2} $
where $k$ is the Coulomb constant (often written as $k = \frac{1}{4\pi\epsilon_0}$). For simplicity, we will keep $k$ as is in the expressions.
Step 4: Calculate Force from the Charge q at x = d/2
The distance between the charge +Q at x = 0 and the charge q at x = d/2 is $d/2$. So the force on +Q (at x=0) due to q is:
$ F_1 = k \frac{Q \, q}{(d/2)^2} = k \frac{Q \, q}{(d^2/4)} = 4k \frac{Q \, q}{d^2} $
Since +Q is at x=0 and q is at x=d/2, if q is positive, the force on +Q is to the right; if q is negative, the force on +Q is to the left.
Step 5: Calculate Force from the Charge +Q at x = d
The distance between the charge +Q at x = 0 and this +Q at x = d is $d$. Hence the force on +Q (at x=0) due to the other +Q at x=d is:
$ F_2 = k \frac{Q \, Q}{d^2} = k \frac{Q^2}{d^2} $
Both charges are positive, so this force is repulsive and acts to the right on the +Q at x=0.
Step 6: Net Force Condition
The problem states that the net force on +Q at x=0 is zero. Labeling rightward as positive, the net force can be written as:
$ F_{\text{net}} = F_1 + F_2 = 0 $
But we must carefully consider the direction. If q is negative, $F_1$ would be attractive and hence directed to the left (negative direction). The force $F_2$ is always to the right. Thus the correct condition is:
$ -\,F_1 + F_2 = 0 $
(since $F_1$ will be in the negative x-direction if q is negative). Substituting the magnitudes:
$ - \left(4k \frac{Q \, q}{d^2}\right) + k \frac{Q^2}{d^2} = 0 $
Step 7: Solve for q
Rearranging the equation:
$ k \frac{Q^2}{d^2} = 4k \frac{Q \, q}{d^2} $
Canceling out common factors $k$, $\frac{1}{d^2}$, and $Q$ (assuming $Q \neq 0$ and $d \neq 0$):
$ Q = 4q $
Thus,
$ q = \frac{Q}{4} $
But to ensure the net force is actually zero in the correct direction, q must be negative (because the force from q at x = d/2 must pull the +Q at x=0 to the left to balance the repulsion from the +Q at x=d). Hence:
$ q = -\,\frac{Q}{4} $
Step 8: Final Answer
The required value of q is $ -\frac{Q}{4} $. This makes the net force on the +Q at x=0 equal to zero.
