The electric field \( E \) is related to the potential difference \( \Delta V \) by:

\[
\Delta V = -\int_{x_1}^{x_2} E \, dx
\]

Given \( E = \frac{100}{x^2} \), the potential difference between \( x = 10 \) m and \( x = 20 \) m is:

\[
\Delta V = -\int_{10}^{20} \frac{100}{x^2} \, dx
\]

\[
\Delta V = -\left[ \frac{100}{x} \right]_{10}^{20}
\]

\[
\Delta V = -\left( \frac{100}{20} - \frac{100}{10} \right)
\]

\[
\Delta V = -\left( 5 - 10 \right) = 5 \, \text{V}
\]

Thus, the potential difference is \( 5 \, \text{V} \).

Electric potential decreases in the direction of Electric Field. So Field will be maximum at A

Work done in a conservative field is not dependent on path taken by the object so,

\[ dV= - \overrightarrow{E}.\overrightarrow{dr}\]

\[ V= - E\widehat{i}.(-a\widehat{i}-b\widehat{j})\]

so, Potential difference = -E.a

Work Done= -Q. E.a

On interchanging A and B with C and D direction of electric field will get reversed. But potential being scaler quantity remains unchanged.

To determine the electric field \(\vec{E}\) from the equipotential surfaces:

1. Electric Field Magnitude: The electric field is the negative gradient of the potential:
\[
|\vec{E}| = -\frac{\Delta V}{\Delta s}
\]
where \(\Delta V\) is the potential difference and \(\Delta s\) is the perpendicular spacing between equipotential surfaces.

2. Horizontal Component (\(E_x\)):
- The horizontal spacing between 0V and 10V surfaces is \(2 \, \text{units}\).
- \(E_x = -\frac{\Delta V_x}{\Delta x} = -\frac{10}{2} = -5 \, \text{V/unit}\).

3. Vertical Component (\(E_y\)):
- The vertical spacing between equipotential surfaces (tilted at \(37^\circ\)) corresponds to \(\tan 37^\circ = \frac{3}{4}\). Hence, the vertical spacing between 0V and 10V is \(2 \cdot \frac{3}{4} = 1.5 \, \text{units}\).
- \(E_y = -\frac{\Delta V_y}{\Delta y} = -\frac{10}{1.5} = -\frac{20}{3} \, \text{V/unit}\).

4. Resultant Electric Field:
\[
\vec{E} = -E_x \hat{i} - E_y \hat{j} = 5\hat{i} + \frac{20}{3}\hat{j}.
\]

Electric Potential (\( V \)):

1. Potential at the center due to a charge \( q \) at a distance \( r \) is:
\[
V = \frac{kq}{r}.
\]

2. Total potential is the algebraic sum since potential is scalar. For 3 positive charges and 5 negative charges:
\[
V_{\text{total}} = 3 \cdot \frac{kq}{r} - 5 \cdot \frac{kq}{r} = \frac{-2kq}{r}.
\]

---

Electric Field (\( E \)):

1. Symmetry of arrangement: The charges along the diagonals cancel their horizontal components, leaving only vertical components.

2. Field due to charges along diagonals (\( -q, -q, +q, +q \)):
- Each diagonal pair creates a net field of magnitude:
\[
E_{\text{pair}} = \frac{\sqrt{2}kq}{r^2}.
\]
- Two such pairs contribute:
\[
E_{\text{diagonals}} = 2 \cdot \frac{\sqrt{2}kq}{r^2} = \frac{2\sqrt{2}kq}{r^2}.
\]

3. Field due to the remaining vertical pair (\( -q, -q \)):
- Net field:
\[
E_{\text{vertical}} = \frac{2kq}{r^2}.
\]

4. Total electric field:
\[
E_{\text{total}} = E_{\text{diagonals}} + E_{\text{vertical}} = \frac{2kq}{r^2}(1 + \sqrt{2}).
\]

---

Final Answer:
\[
E = \frac{2kq}{r^2}(1 + \sqrt{2}), \quad V = \frac{-2kq}{r}.
\]

The electric field \(\vec{E}\) is related to the potential \(V\) by:
\[
\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}\right).
\]

Given \(V = k[2x^2 + y^2 - 2z]\):

1. Partial derivatives:
- \(\frac{\partial V}{\partial x} = 4kx\),
- \(\frac{\partial V}{\partial y} = 2ky\),
- \(\frac{\partial V}{\partial z} = -2k\).

2. Electric field components at \((1, 0, 1)\):
- \(E_x = -\frac{\partial V}{\partial x} = -4k(1) = -4k\),
- \(E_y = -\frac{\partial V}{\partial y} = -2k(0) = 0\),
- \(E_z = -\frac{\partial V}{\partial z} = -(-2k) = 2k\).

3. Magnitude of \(\vec{E}\):
\[
|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-4k)^2 + (0)^2 + (2k)^2} = \sqrt{16k^2 + 4k^2} = \sqrt{20k^2} = 2k\sqrt{5}.
\]

Thus, the magnitude of the electric field is \(2k\sqrt{5}\).

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

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

1. Distance from the charge at the origin:
- Point \( A(\sqrt{2}, \sqrt{2}) \): Distance \( r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2 \).
- Point \( B(2, 0) \): Distance \( r_B = \sqrt{(2)^2 + 0^2} = 2 \).

2. Potentials at \( A \) and \( B \):
\[
V_A = \frac{kq}{r_A} = \frac{kq}{2}, \quad V_B = \frac{kq}{r_B} = \frac{kq}{2}.
\]

3. Potential difference:
\[
V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0.
\]

Thus, \( V_A - V_B = 0 \).