Electric charges of `+10 muC,+5 muC,-3 muC and +8 muC` are placed at the corners of a square of side `sqrt(2)` m, the potential at the centre of the square is

Options

A. 1.8 V
B. `1.8 xx 10^(6)V`
C. `1.8 xx 10^(5)V`
D. `1.8 xx 10^(4)V`

Answer

To find the electric potential at the center of a square with given charges at its corners, we can follow these steps:

### Step 1: Identify the charges and their positions
We have four charges located at the corners of a square:
- Charge \( Q_A = +10 \, \mu C \) at corner A
- Charge \( Q_B = +5 \, \mu C \) at corner B
- Charge \( Q_C = -3 \, \mu C \) at corner C
- Charge \( Q_D = +8 \, \mu C \) at corner D

### Step 2: Calculate the distance from the center to each corner
The side length of the square is \( \sqrt{2} \, m \). The distance from the center of the square to any corner can be calculated using the Pythagorean theorem. The diagonal \( d \) of the square is given by:
\[ d = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, m \]
Since the center divides the diagonal into two equal parts, the distance from the center to each corner is:
\[ r = \frac{d}{2} = \frac{2}{2} = 1 \, m \]

### Step 3: Use the formula for electric potential
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \frac{Q}{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step 4: Calculate the potential at the center due to each charge
The total potential at the center \( V_O \) is the algebraic sum of the potentials due to each charge:
\[ V_O = V_A + V_B + V_C + V_D \]
Calculating each term:
- For \( Q_A \):
\[ V_A = k \frac{Q_A}{r} = 9 \times 10^9 \frac{10 \times 10^{-6}}{1} = 90,000 \, V \]
- For \( Q_B \):
\[ V_B = k \frac{Q_B}{r} = 9 \times 10^9 \frac{5 \times 10^{-6}}{1} = 45,000 \, V \]
- For \( Q_C \):
\[ V_C = k \frac{Q_C}{r} = 9 \times 10^9 \frac{-3 \times 10^{-6}}{1} = -27,000 \, V \]
- For \( Q_D \):
\[ V_D = k \frac{Q_D}{r} = 9 \times 10^9 \frac{8 \times 10^{-6}}{1} = 72,000 \, V \]

### Step 5: Sum the potentials
Now, we sum all the potentials:
\[ V_O = V_A + V_B + V_C + V_D = 90,000 + 45,000 - 27,000 + 72,000 \]
Calculating this gives:
\[ V_O = 90,000 + 45,000 + 72,000 - 27,000 = 180,000 \, V \]

### Step 6: Convert to scientific notation
\[ V_O = 1.8 \times 10^5 \, V \]

### Final Answer
The potential at the center of the square is \( 1.8 \times 10^5 \, V \).

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