ABCD is a square of side `0.2m`. Charges of `2xx10^(-9) C, 4xx10^(-9) C and 8xx10^(-9) C` are placed at the corners A, B and C respectively. Calculate work required to transfer a charge of `2xx10^(-9) C` from corner D to centre of the square.

Explanation

To solve the problem of calculating the work required to transfer a charge of \(2 \times 10^{-9} \, C\) from corner D to the center of the square ABCD, we will follow these steps:

### Step 1: Determine the potential at the center of the square (O)
The potential at a point due to a point charge is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(Q\) is the charge, and \(r\) is the distance from the charge to the point where we are calculating the potential.

For the square ABCD with side \(0.2 \, m\):
- The distance from each corner (A, B, C) to the center (O) can be calculated using the Pythagorean theorem. Since the center divides the square into two equal halves, the distance \(r\) from any corner to the center is:
\[ r = \frac{0.2}{\sqrt{2}} = 0.1\sqrt{2} \, m \]

Now, we calculate the potential at the center (O) due to each charge:
1. Charge at A: \(Q_1 = 2 \times 10^{-9} \, C\)
\[ V_A = \frac{k \cdot Q_1}{r} = \frac{9 \times 10^9 \cdot 2 \times 10^{-9}}{0.1\sqrt{2}} = \frac{18 \times 10^0}{0.1\sqrt{2}} = \frac{180}{\sqrt{2}} \, V \]

2. Charge at B: \(Q_2 = 4 \times 10^{-9} \, C\)
\[ V_B = \frac{k \cdot Q_2}{r} = \frac{9 \times 10^9 \cdot 4 \times 10^{-9}}{0.1\sqrt{2}} = \frac{36 \times 10^0}{0.1\sqrt{2}} = \frac{360}{\sqrt{2}} \, V \]

3. Charge at C: \(Q_3 = 8 \times 10^{-9} \, C\)
\[ V_C = \frac{k \cdot Q_3}{r} = \frac{9 \times 10^9 \cdot 8 \times 10^{-9}}{0.1\sqrt{2}} = \frac{72 \times 10^0}{0.1\sqrt{2}} = \frac{720}{\sqrt{2}} \, V \]

Now, we sum the potentials to find the total potential at the center (O):
\[ V_O = V_A + V_B + V_C = \frac{180 + 360 + 720}{\sqrt{2}} = \frac{1260}{\sqrt{2}} \approx 891.1 \, V \]

### Step 2: Determine the potential at corner D
The potential at corner D is calculated similarly, considering the distances to each charge:
1. Charge at A:
\[ V_A = \frac{k \cdot Q_1}{0.2} = \frac{9 \times 10^9 \cdot 2 \times 10^{-9}}{0.2} = 90 \, V \]

2. Charge at B:
\[ V_B = \frac{k \cdot Q_2}{0.2\sqrt{2}} = \frac{9 \times 10^9 \cdot 4 \times 10^{-9}}{0.2\sqrt{2}} = \frac{180}{\sqrt{2}} \approx 127.3 \, V \]

3. Charge at C:
\[ V_C = \frac{k \cdot Q_3}{0.2} = \frac{9 \times 10^9 \cdot 8 \times 10^{-9}}{0.2} = 360 \, V \]

Now, we sum the potentials to find the total potential at D:
\[ V_D = V_A + V_B + V_C = 90 + 127.3 + 360 \approx 577.3 \, V \]

### Step 3: Calculate the work done to transfer the charge
The work done \(W\) in moving a charge \(q\) from a point with potential \(V_D\) to a point with potential \(V_O\) is given by:
\[ W = q \cdot (V_O - V_D) \]
Substituting the values:
\[ W = 2 \times 10^{-9} \cdot (891.1 - 577.3) = 2 \times 10^{-9} \cdot 313.8 \approx 6.276 \times 10^{-7} \, J \]

### Final Answer
The work required to transfer a charge of \(2 \times 10^{-9} \, C\) from corner D to the center of the square is approximately \(6.276 \times 10^{-7} \, J\).

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