Solution:

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 \( 5 \, \text{V} \).

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Solution:

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

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Solution:

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

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Solution:

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

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Solution:

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

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Solution:

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:

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

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Solution:

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:

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

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Solution:

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.

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Solution:

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, \( n = 6 \).

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Solution:

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 \( Q = -q \).

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