\[ E=-\frac{d V}{dr}= -(\frac{-5}{2})= 2.5 V/m \]

The electric potential \( V \) at the center of a charged half-ring is given by:

\[
V = \frac{1}{4\pi \epsilon_0} \int \frac{\lambda \, dl}{r}
\]

For a half-ring of radius \( r \), the total length is \( \pi r \), and the potential at the center is:

\[
V = \frac{\lambda}{4\pi \epsilon_0} \cdot \pi r \cdot \frac{1}{r} = \frac{\lambda}{4\epsilon_0}
\]

Thus, the electric potential at the center of the half-ring is:

\[
V = \frac{\lambda}{4 \epsilon_0}
\]

The electric field \(\vec{E}\) is related to the electric potential \(V\) by:

\[
\vec{E} = -\nabla V
\]

Given \(V = -x^2 y - xz^3 + 4\), we calculate the gradient of \(V\):

\[
E_x = -\frac{\partial V}{\partial x} = 2xy + z^3
\]
\[
E_y = -\frac{\partial V}{\partial y} = x^2
\]
\[
E_z = -\frac{\partial V}{\partial z} = 3xz^2
\]

Thus, the electric field is:

\[
\vec{E} = i(2xy + z^3) + j(x^2) + k(3xz^2)
\]

For a solid, non-conducting sphere uniformly charged throughout its volume, the electric potential \( V(r) \) is determined as follows:

---

Inside the sphere (\( r \leq R \)):
Using Gauss's law, the electric field inside the sphere is:
\[
E = \frac{\rho r}{3\epsilon_0},
\]
where \( \rho \) is the charge density, and \( R \) is the sphere's radius.

The potential \( V(r) \) at a distance \( r \) from the center is obtained by integrating the electric field:
\[
V(r) = V(R) - \int_r^R E \, dr = V(R) - \int_r^R \frac{\rho r}{3\epsilon_0} \, dr.
\]
Simplifying:
\[
V(r) = V(R) - \frac{\rho}{6\epsilon_0} \left( R^2 - r^2 \right).
\]
Thus, inside the sphere, \( V(r) \) is a quadratic function of \( r \), decreasing as \( r^2 \).

---

Outside the sphere (\( r > R \)):
The sphere acts as a point charge with total charge \( Q = \frac{4}{3} \pi R^3 \rho \). The potential outside is:
\[
V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}.
\]
Here, \( V(r) \) decreases inversely with \( r \).

Behavior of the Graph:
1. At \( r = R \): The potential is continuous, with its value matching between the inside and outside solutions.
2. Inside the sphere (\( r < R \)): \( V(r) \) decreases quadratically as \( r^2 \).
3. Outside the sphere (\( r > R \)): \( V(r) \) decreases inversely with \( r \).

Thus, the graph shows a continuous potential that decreases quadratically inside the sphere and decreases inversely outside the sphere, matching the provided curve.

To find the charge enclosed, we use Gauss's law:

\[
Q = \oint \vec{E} \cdot d\vec{A} = \epsilon_0 \frac{dV}{dx} \cdot A
\]

Given \( V(x) = 3 - 2x^3 \), we compute the electric field:

\[
E_x = -\frac{dV}{dx} = 6x^2
\]

Now, for a cube with side 1m and one vertex at the origin, the flux through the cube's face on the x-axis is:

\[
\text{Flux} = E_x \cdot A = 6x^2 \cdot 1 = 6
\]

Since only the face along the x-axis contributes to the flux, the total charge enclosed is:

\[
Q = \epsilon_0 \cdot 6
\]

Thus, \( n = 6 \).

The potential at the center of the square is the sum of the potentials due to each charge. Since potential is a scalar quantity, the total potential at the center is the algebraic sum of individual potentials.

For the potential to be zero, the sum of the potentials must cancel out. The relationship between \( Q \) and \( q \) for this condition is:

\[
\frac{-Q}{r} + \frac{-q}{r} + \frac{2q}{r} + \frac{2Q}{r} = 0
\]

Simplifying:

\[
(-Q + 2Q) + (-q + 2q) = 0
\]

\[
Q = -q
\]

Thus, the relation is \( Q = -q \).

The graph shown correctly represents the variation of electric potential along the x-axis between two negative charges.

Explanation:
1. Electric Potential Due to a Point Charge:
- For a single negative charge, the potential is negative and becomes less negative (closer to zero) as the distance from the charge increases.

2. Superposition of Potentials:
- The net potential at any point is the sum of potentials due to both charges.
- Near each charge, the potential is dominated by that charge and highly negative.
- At a point midway between the two charges, the potentials due to both charges add up (both are negative), giving the minimum (most negative) potential.

3. Shape of the Graph:

- The potential decreases sharply near each charge.
- Between the charges, the potential reaches a minimum (most negative value) due to the additive contributions of both charges.

The graph accurately shows this behavior with a dip between the two charges.

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

For a uniformly charged ring:

1. Electric potential on the axis:
\[
V = \frac{kQ}{\sqrt{R^2 + x^2}},
\]
where \( R = \sqrt{5} \) and \( x \) is the distance from the center.

2. Electric field on the axis:
\[
E = \frac{kQx}{(R^2 + x^2)^{3/2}}.
\]

3. Given:
\( V = 6 \, \text{V} \), \( E = 1 \, \text{V/m} \).

4. Divide \( E \) by \( V \) to eliminate \( kQ \):
\[
\frac{E}{V} = \frac{x}{R^2 + x^2} \quad \Rightarrow \quad \frac{1}{6} = \frac{x}{5 + x^2}.
\]

5. Simplify:
\[
5 + x^2 = 6x \quad \Rightarrow \quad x^2 - 6x + 5 = 0.
\]

6. Solve quadratic:
\[
(x - 5)(x - 1) = 0 \quad \Rightarrow \quad x = 5 \, \text{or} \, x = 1.
\]

The smallest value of \( x \) is 1 m.

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

Given \( V = \frac{1}{2}(y^2 - 4x) \):

1. Partial derivatives:
- \(\frac{\partial V}{\partial x} = -2\)
- \(\frac{\partial V}{\partial y} = y\).

2. At \( x = 1, y = 1 \):
- \( E_x = -\frac{\partial V}{\partial x} = -(-2) = 2 \),
- \( E_y = -\frac{\partial V}{\partial y} = -(1) = -1 \).

3. Electric field:
\[
\vec{E} = 2\hat{i} - \hat{j} \, \text{V/m}.
\]