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 find the value of the integral

\[ -\int_{-\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} \]

given that

\[ -\int_{\frac{a}{\sqrt{2}}}^{0} \vec{E} \cdot d\vec{r} = \frac{\sqrt{2}Kq}{a} \]

where \( K = \frac{1}{4\pi \epsilon_0} \).

### Step-by-step Solution:

1. **Understanding the Electric Field**:
The electric field \( \vec{E} \) due to a point charge \( +q \) at a distance \( r \) is given by:
\[ \vec{E} = K \frac{q}{r^2} \hat{r} \]
where \( K = \frac{1}{4\pi \epsilon_0} \).

2. **Setting Up the Integral**:
We need to evaluate the integral from \(-\infty\) to \(\frac{a}{\sqrt{2}}\). We can break this into two parts:
\[ -\int_{-\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} = -\int_{-\infty}^{0} \vec{E} \cdot d\vec{r} - \int_{0}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} \]

3. **Using the Given Value**:
We know from the problem statement that:
\[ -\int_{\frac{a}{\sqrt{2}}}^{0} \vec{E} \cdot d\vec{r} = \frac{\sqrt{2}Kq}{a} \]
Therefore, we can substitute this into our expression:
\[ -\int_{-\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} = -\left(-\int_{0}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} + \frac{\sqrt{2}Kq}{a}\right) \]

4. **Combining the Integrals**:
The integral from \( 0 \) to \( \frac{a}{\sqrt{2}} \) is the remaining part. Let's denote it as \( V_0 \):
\[ V_0 = -\int_{0}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} \]
Thus, we have:
\[ -\int_{-\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} = V_0 + \frac{\sqrt{2}Kq}{a} \]

5. **Finding the Value of the Integral**:
Now, we need to find the total potential difference. The potential at infinity is zero, and at \( \frac{a}{\sqrt{2}} \) we have:
\[ V = 4Kq \frac{\sqrt{2}}{a} \]
Therefore, we can express the total integral as:
\[ V - 0 = 4Kq \frac{\sqrt{2}}{a} - \frac{\sqrt{2}Kq}{a} \]
Simplifying gives:
\[ V = 3Kq \frac{\sqrt{2}}{a} \]

6. **Final Answer**:
Thus, the value of the integral is:
\[ -\int_{-\infty}^{\frac{a}{\sqrt{2}}} \vec{E} \cdot d\vec{r} = 3Kq \frac{\sqrt{2}}{a} \]