Two charges q and -q are placedat ( a,0) and `( - a,0)` on the x-axis. Another charge 2q is taken from ( 0,0) to ( 0,a) , then the work done to do so

Options

A. `(q)/( 2 pi epsilon_(0)a)`
B. Zero
C. Infinite
D. `(q)/( 4 pi epsilon_(0)a)`

Answer

To solve the problem of calculating the work done in moving a charge \(2q\) from the origin \((0,0)\) to the point \((0,a)\) in the presence of two charges \(q\) and \(-q\) located at \((a,0)\) and \((-a,0)\) respectively, we can follow these steps:

### Step 1: Determine the Electric Potential at the Points of Interest
The electric potential \(V\) due to a point charge \(Q\) at a distance \(r\) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \(k\) is Coulomb's constant.

### Step 2: Calculate the Potential at the Origin (0,0)
For the charge \(q\) at \((a,0)\):
- The distance from \((0,0)\) to \((a,0)\) is \(a\).
- The potential due to charge \(q\) at the origin is:
\[ V_1 = \frac{kq}{a} \]

For the charge \(-q\) at \((-a,0)\):
- The distance from \((0,0)\) to \((-a,0)\) is also \(a\).
- The potential due to charge \(-q\) at the origin is:
\[ V_2 = \frac{k(-q)}{a} = -\frac{kq}{a} \]

Now, the total potential at the origin \(V_O\) is:
\[ V_O = V_1 + V_2 = \frac{kq}{a} - \frac{kq}{a} = 0 \]

### Step 3: Calculate the Potential at the Point (0,a)
Now, we calculate the potential at the point \((0,a)\).

For the charge \(q\) at \((a,0)\):
- The distance from \((0,a)\) to \((a,0)\) is given by:
\[ r_1 = \sqrt{(a-0)^2 + (0-a)^2} = \sqrt{a^2 + a^2} = a\sqrt{2} \]
- The potential due to charge \(q\) at \((0,a)\) is:
\[ V_1' = \frac{kq}{a\sqrt{2}} \]

For the charge \(-q\) at \((-a,0)\):
- The distance from \((0,a)\) to \((-a,0)\) is also:
\[ r_2 = \sqrt{(-a-0)^2 + (0-a)^2} = \sqrt{a^2 + a^2} = a\sqrt{2} \]
- The potential due to charge \(-q\) at \((0,a)\) is:
\[ V_2' = \frac{k(-q)}{a\sqrt{2}} = -\frac{kq}{a\sqrt{2}} \]

Now, the total potential at the point \((0,a)\) is:
\[ V_P = V_1' + V_2' = \frac{kq}{a\sqrt{2}} - \frac{kq}{a\sqrt{2}} = 0 \]

### Step 4: Calculate the Work Done
The work done \(W\) in moving the charge \(2q\) from the origin to the point \((0,a)\) is given by:
\[ W = q(V_P - V_O) \]
Substituting the values we found:
\[ W = 2q(0 - 0) = 0 \]

### Final Answer
The work done in moving the charge \(2q\) from \((0,0)\) to \((0,a)\) is:
\[ \boxed{0} \]