To find the coordinates where the net electric field is zero and where it is maximum for two identical charges \( Q \) placed at coordinates \((-a, 0)\) and \((a, 0)\), we can follow these steps:
### Step 1: Understanding the Electric Field due to Point Charges
The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[
E = \frac{kQ}{r^2}
\]
where \( k \) is Coulomb's constant.
### Step 2: Analyzing the Electric Field at the Origin
For the two identical charges located at \((-a, 0)\) and \((a, 0)\):
- The electric field at the origin (0,0) due to the charge at \((-a, 0)\) is directed to the right (positive x-direction).
- The electric field at the origin due to the charge at \((a, 0)\) is directed to the left (negative x-direction).
Since both fields have the same magnitude and are equal in magnitude but opposite in direction, they cancel each other out at the origin. Therefore, the net electric field is zero at the origin.
### Step 3: Finding the Point where the Electric Field is Maximum
To find the points where the electric field is maximum, we need to consider points along the y-axis, as the electric field will be symmetric about the x-axis.
Let’s denote a point \( P \) at coordinates \( (0, y) \).
The distance from the charge at \((-a, 0)\) to point \( P \) is:
\[
d_1 = \sqrt{a^2 + y^2}
\]
The distance from the charge at \((a, 0)\) to point \( P \) is:
\[
d_2 = \sqrt{a^2 + y^2}
\]
The electric fields due to both charges at point \( P \) will have the same magnitude but will point away from the charges.
### Step 4: Expression for the Net Electric Field
The electric field at point \( P \) due to both charges can be expressed as:
\[
E_{net} = E_1 + E_2 = \frac{kQ}{d_1^2} + \frac{kQ}{d_2^2}
\]
Since \( d_1 = d_2 \):
\[
E_{net} = 2 \cdot \frac{kQ}{(a^2 + y^2)}
\]
### Step 5: Finding the Maximum Electric Field
To find the maximum electric field, we differentiate \( E_{net} \) with respect to \( y \) and set the derivative equal to zero:
\[
\frac{dE_{net}}{dy} = 0
\]
Calculating the derivative, we find:
\[
\frac{d}{dy}\left( \frac{2kQ}{a^2 + y^2} \right) = 0
\]
This leads to the condition:
\[
y^2 = \frac{a^2}{2}
\]
Thus, the coordinates where the electric field is maximum are:
\[
y = \pm \frac{a}{\sqrt{2}}
\]
So the points are:
\[
(0, \frac{a}{\sqrt{2}}) \text{ and } (0, -\frac{a}{\sqrt{2}})
\]
### Final Answer
- The point where the net electric field is zero: \( (0, 0) \)
- The points where the electric field is maximum: \( (0, \frac{a}{\sqrt{2}}) \) and \( (0, -\frac{a}{\sqrt{2}}) \)
