Two positive point charges of magnitude Q each are separated by a distance "2a". A test charge `'g_0'` is located in a plane which is normal to the line joining these charges and midway between them. The locus of the points in this plane for which the force on the test charge has a maximum value is

Options

A. a circle of radius `r=a//sqrt2`
B. a circle of radius r=a
C. a straight line
D. an ellipse

Answer

To solve the problem, we need to find the locus of points in a plane where the force on a test charge \( g_0 \) is maximized due to two positive point charges \( Q \) separated by a distance \( 2a \). Here’s the step-by-step solution:

### Step 1: Understand the Configuration
We have two positive point charges \( Q \) located at points \( A \) and \( B \) with a distance of \( 2a \) between them. The test charge \( g_0 \) is placed in a plane that is perpendicular to the line joining \( A \) and \( B \) and is equidistant from both charges.

### Step 2: Define the Geometry
Let’s place charge \( Q \) at \( A(-a, 0) \) and \( B(a, 0) \). The test charge \( g_0 \) is located at a point \( (0, r) \) in the plane, where \( r \) is the distance from the midpoint of \( AB \) to the test charge.

### Step 3: Calculate the Forces
The electrostatic force \( F \) on the test charge \( g_0 \) due to each charge \( Q \) can be calculated using Coulomb's law:
\[ F = k \frac{Q g_0}{r_{AB}^2} \]
where \( r_{AB} \) is the distance from \( g_0 \) to either charge \( Q \). By the Pythagorean theorem:
\[ r_{AB} = \sqrt{a^2 + r^2} \]
Thus, the force due to each charge is:
\[ F = k \frac{Q g_0}{(a^2 + r^2)} \]

### Step 4: Resolve Forces into Components
Due to symmetry, the horizontal components of the forces due to both charges will cancel each other out, while the vertical components will add up. The vertical component of the force from each charge is:
\[ F_y = F \sin \theta \]
where \( \sin \theta = \frac{r}{\sqrt{a^2 + r^2}} \). Therefore, the net vertical force \( F_{net} \) on \( g_0 \) is:
\[ F_{net} = 2 F \sin \theta = 2 \left(k \frac{Q g_0}{(a^2 + r^2)}\right) \left(\frac{r}{\sqrt{a^2 + r^2}}\right) \]
This simplifies to:
\[ F_{net} = 2kQ g_0 \frac{r}{(a^2 + r^2)^{3/2}} \]

### Step 5: Maximize the Force
To find the maximum force, we need to maximize \( F_{net} \) with respect to \( r \). We can treat \( F_{net} \) as a function of \( r \):
\[ F_{net}(r) = 2kQ g_0 \frac{r}{(a^2 + r^2)^{3/2}} \]
To find the maximum, we take the derivative \( \frac{dF_{net}}{dr} \) and set it to zero:
\[ \frac{dF_{net}}{dr} = 0 \]

### Step 6: Solve the Derivative
After differentiating and simplifying, we set the equation:
\[ (a^2 + r^2)^{3/2} - 3r^2(a^2 + r^2)^{1/2} = 0 \]
Factoring out \( (a^2 + r^2)^{1/2} \) gives:
\[ (a^2 + r^2) - 3r^2 = 0 \]
This leads to:
\[ a^2 - 2r^2 = 0 \implies r^2 = \frac{a^2}{2} \implies r = \frac{a}{\sqrt{2}} \]

### Step 7: Determine the Locus
The locus of points where the force on the test charge \( g_0 \) is maximized is a circle of radius \( r = \frac{a}{\sqrt{2}} \) centered at the midpoint between the two charges in the plane.

### Conclusion
The locus of points in the plane where the force on the test charge has a maximum value is a circle of radius \( \frac{a}{\sqrt{2}} \).

---