Solution:

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.

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Solution:

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

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Solution:

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:
\[
V_{\text{total}} = 3 \cdot \frac{kq}{r} - 5 \cdot \frac{kq}{r} = \frac{-2kq}{r}.
\]

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

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

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Solution:

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.

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Solution:

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

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Solution:

As potential is a scaler quantity distribution of charges will not have any effect.

Total charge = 3Q-2Q+Q= 2Q

Electric Potential V = \( \frac{2KQ}{R}= \frac{Q}{2\pi \varepsilon_{0}R}\)

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Solution:

In a uniform electric field, the electric potential decreases in the direction of the field. From the diagram:

1. Direction of the electric field: The field points downward and to the left (parallel to lines \(C \to D\)).

2. Potential relation:
- Points further along the direction of the field have lower potential.
- Points opposite to the field direction have higher potential.

3. Comparison of potentials:
- \(A\): Least aligned with the field direction (highest potential).
- \(B\): Slightly more aligned with the field direction than \(A\).
- \(D\): More aligned with the field direction than \(B\).
- \(C\): Most aligned with the field direction (lowest potential).

Thus, the potential order is:
\[
V_A > V_B > V_D > V_C
\]

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Solution:

At the centre of a uniformly charge ring electric field becomes zero because being a vector quantity it will cancel out each.

Electric potential is a scaler quantity so will be present at the centre of ring.

E = 0, V = maximum

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Solution:

The electric potential \( V \) at the center of the cube due to a point charge \( q \) is:
\[
V = \frac{kq}{r},
\]
where \( r \) is the distance of the charge from the center.

1. Distance from center:
For a cube of side \( b \), the distance of any vertex from the center is:
\[
r = \frac{\sqrt{3}b}{2}.
\]

2. Potential due to 7 charges:
Since potential is scalar, the total potential is the sum of potentials due to all charges:
\[
V_{\text{total}} = 7 \cdot \frac{kq}{r}.
\]

Substitute \( r = \frac{\sqrt{3}b}{2} \):
\[
V_{\text{total}} = 7 \cdot \frac{kq}{\frac{\sqrt{3}b}{2}} = 7 \cdot \frac{2kq}{\sqrt{3}b} = \frac{14kq}{\sqrt{3}b}.
\]

Thus, the potential at the center is \(\frac{14kq}{\sqrt{3}b}\).

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Solution:

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 \(2k\sqrt{5}\).

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