Equal charges q are placed at the three corners B, C, D of a square ABCD of side 'a'. The ratio of potential at A to centre O is

Options

A. `(2sqrt(2)+1)/(3sqrt(2)) `
B. `(2+sqrt(2))/(6)`
C. `(2sqrt(2)+1)/(6)`
D. `(2sqrt(2)+1)/(3)`

Explanation

To solve the problem of finding the ratio of the electric potential at point A to that at the center O of the square ABCD, we will follow these steps:

### Step 1: Calculate the potential at point A

The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by the formula:

\[ V = \frac{k \cdot q}{r} \]

where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

In our case, we have three charges \( q \) located at points B, C, and D. We need to find the distance from point A to each of these charges:

1. Distance from A to B: \( AB = a \)
2. Distance from A to C: \( AC = \sqrt{a^2 + a^2} = a\sqrt{2} \)
3. Distance from A to D: \( AD = a \)

Now, we can calculate the total potential at point A due to the three charges:

\[ V_A = V_B + V_C + V_D \]

Substituting the distances into the potential formula:

\[ V_A = \frac{k \cdot q}{a} + \frac{k \cdot q}{a\sqrt{2}} + \frac{k \cdot q}{a} \]

Combining the terms:

\[ V_A = \frac{k \cdot q}{a} + \frac{k \cdot q}{a} + \frac{k \cdot q}{a\sqrt{2}} = \frac{2kq}{a} + \frac{kq}{a\sqrt{2}} \]

### Step 2: Calculate the potential at the center O

The center O of the square is at a distance of \( \frac{a}{\sqrt{2}} \) from each of the corners B, C, and D. Therefore, the potential at point O due to each charge is:

\[ V_O = V_B + V_C + V_D \]

Calculating the potential at O:

\[ V_O = \frac{k \cdot q}{\frac{a}{\sqrt{2}}} + \frac{k \cdot q}{\frac{a}{\sqrt{2}}} + \frac{k \cdot q}{\frac{a}{\sqrt{2}}} \]

This simplifies to:

\[ V_O = 3 \cdot \frac{k \cdot q \cdot \sqrt{2}}{a} \]

### Step 3: Find the ratio of potentials \( \frac{V_A}{V_O} \)

Now we can find the ratio of the potentials at A and O:

\[ \frac{V_A}{V_O} = \frac{\left( \frac{2kq}{a} + \frac{kq}{a\sqrt{2}} \right)}{3 \cdot \frac{kq \cdot \sqrt{2}}{a}} \]

Canceling \( kq/a \):

\[ \frac{V_A}{V_O} = \frac{2 + \frac{1}{\sqrt{2}}}{3\sqrt{2}} \]

### Step 4: Simplify the ratio

To simplify further, we can express \( \frac{1}{\sqrt{2}} \) as \( \frac{\sqrt{2}}{2} \):

\[ \frac{V_A}{V_O} = \frac{2 + \frac{\sqrt{2}}{2}}{3\sqrt{2}} = \frac{4 + \sqrt{2}}{6\sqrt{2}} \]

Thus, the final ratio of the potential at A to that at O is:

\[ \frac{V_A}{V_O} = \frac{4 + \sqrt{2}}{6\sqrt{2}} \]