Four idenbtial charges `+50 mu C` each are placed, one at each corner of a square of side `2 m`. How much external energy is required to bring another charge of `+50 mu C` from infinity to the centre of the square
(Given `(1)/(4pi epsilon_(0)) = 9 xx 10^(9)(Nm^(2))/(C^(2)))`

Options

A. 64 J
B. 41 J
C. 16 J
D. 10 J

Answer

To solve the problem of how much external energy is required to bring another charge of \( +50 \, \mu C \) from infinity to the center of a square with four identical charges of \( +50 \, \mu C \) each placed at its corners, we can follow these steps:

### Step 1: Determine the distance from the center to a corner of the square.
The side length of the square is given as \( 2 \, m \). The distance from the center of the square to any corner can be found using the Pythagorean theorem. The distance \( r \) from the center to a corner is given by:

\[ r = \frac{\sqrt{2}}{2} \times \text{side length} = \frac{\sqrt{2}}{2} \times 2 = \sqrt{2} \, m \]

### Step 2: Calculate the potential at the center due to the four charges.
The electric potential \( V \) at a point due to a point charge is given by:

\[ V = \frac{kQ}{r} \]

where \( k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q \) is the charge, and \( r \) is the distance from the charge to the point.

Since there are four identical charges \( Q = 50 \, \mu C = 50 \times 10^{-6} \, C \) at the corners, the total potential \( V_O \) at the center \( O \) is:

\[ V_O = 4 \times \frac{kQ}{r} \]

Substituting the values:

\[ V_O = 4 \times \frac{9 \times 10^9 \times 50 \times 10^{-6}}{\sqrt{2}} \]

### Step 3: Calculate the work done in bringing the charge from infinity to the center.
The work done \( W \) in bringing a charge \( q \) from infinity to a point where the potential is \( V \) is given by:

\[ W = qV \]

Here, \( q = 50 \, \mu C = 50 \times 10^{-6} \, C \). Thus, substituting the expression for \( V_O \):

\[ W = 50 \times 10^{-6} \times V_O \]

Substituting \( V_O \) from Step 2:

\[ W = 50 \times 10^{-6} \times \left( 4 \times \frac{9 \times 10^9 \times 50 \times 10^{-6}}{\sqrt{2}} \right) \]

### Step 4: Simplify and calculate the final value.
Now we can simplify the expression:

\[ W = 50 \times 10^{-6} \times 4 \times \frac{9 \times 10^9 \times 50 \times 10^{-6}}{\sqrt{2}} \]

Calculating the numerical values:

1. \( 50 \times 50 = 2500 \)
2. \( 4 \times 9 = 36 \)
3. Thus, \( W = \frac{36 \times 2500 \times 10^{-12} \times 10^9}{\sqrt{2}} \)
4. \( = \frac{90000 \times 10^{-3}}{\sqrt{2}} \)
5. \( = \frac{90000}{1.414} \approx 63660.2 \, J \)

So, rounding off, we get:

\[ W \approx 64 \, J \]

### Final Answer:
The external energy required to bring the charge from infinity to the center of the square is approximately \( 64 \, J \).

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