Four similar charges each of charge `+q` are placed at four corners of a square of side a. Find the value of integral `-int_(infty)^(a//sqrt(2)) vec(E).vec(d)r.` (in volts) if value of integral `-int_(a//sqrt(2))^(0) vec(E).vec(dr)=(sqrt(2)Kq)/(a)("where" K=(1)/(4 pi epsilon_(0)))`

Options

A. `(4Kq)/(a)`
B. `(3Kq)/(a)`
C. `(3sqrt(2)Kq)/(a)`
D. `(4sqrt(2)Kq)/(a)`

Answer

To solve the problem, we need to calculate the integral of the electric field \(\vec{E}\) from infinity to \(\frac{a}{\sqrt{2}}\) and relate it to the given integral from \(\frac{a}{\sqrt{2}}\) to \(0\).

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have four charges \(+q\) located at the corners of a square with side length \(a\). The electric field \(\vec{E}\) at any point in space due to these charges can be calculated using the principle of superposition.

2. **Electric Potential Due to Charges**:
The electric potential \(V\) at a point due to a point charge is given by:
\[ V = K \cdot \frac{q}{r} \]
where \(K = \frac{1}{4\pi\epsilon_0}\) and \(r\) is the distance from the charge to the point.

3. **Finding the Electric Field**:
The electric field \(\vec{E}\) is related to the electric potential \(V\) by:
\[ \vec{E} = -\nabla V \]
To find the electric field at a specific point, we can use the contributions from all four charges.

4. **Setting Up the Integral**:
We need to evaluate the integral:
\[ -\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} \]
We can express this integral in terms of the known integral:
\[ -\int_{\frac{a}{\sqrt{2}}}^{0} \vec{E} \cdot d\vec{r} = \frac{\sqrt{2}Kq}{a} \]

5. **Relating the Integrals**:
We know from the properties of electric potential that:
\[ V_{\infty} - V_{\frac{a}{\sqrt{2}}} = -\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} \]
and
\[ V_{\frac{a}{\sqrt{2}}} - V_{0} = -\int_{\frac{a}{\sqrt{2}}}^{0} \vec{E} \cdot d\vec{r} \]
Therefore, we can express the total change in potential from infinity to zero as:
\[ V_{\infty} - V_{0} = -\int_{\infty}^{0} \vec{E} \cdot d\vec{r} \]

6. **Calculating the Final Value**:
Since \(V_{0} = 0\) (at the origin), we have:
\[ V_{\infty} = -\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} + \frac{\sqrt{2}Kq}{a} \]
Assuming \(V_{\infty} = 0\) (as we are at infinity), we can simplify:
\[ 0 = -\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} + \frac{\sqrt{2}Kq}{a} \]
Thus, we find:
\[ -\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} = -\frac{\sqrt{2}Kq}{a} \]

### Final Answer:
The value of the integral \(-\int_{\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r}\) is:
\[ -\frac{\sqrt{2}Kq}{a} \]