Q.A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed

Concept & Intuition

Gauss’s law states that the total electric flux through any closed surface equals the charge enclosed divided by $\varepsilon_0$. But “total flux” means the net flux — the sum of contributions over the entire surface, taking direction into account.

• Flux leaving the surface (outward) is taken as positive.
• Flux entering the surface (inward) is taken as negative. So if $\phi_A$ is the outward (leaving) flux and $\phi_B$ is the inward (entering) flux, the net flux is $\phi_A - \phi_B$ (since entering flux contributes negatively). Gauss’s law then gives the enclosed charge directly.

Step-by-step reasoning

1. Recall Gauss’s law For any closed Gaussian surface,

where $\Phi_{\text{net}}$ is the net electric flux through the surface.

2. Interpret the given fluxes

   • $\phi_A$ = flux leaving the surface (outward direction).
   • $\phi_B$ = flux entering the surface (inward direction). By convention, outward flux is positive, inward flux is negative.

3. Write the net flux

   The net flux is the algebraic sum:

4. Apply Gauss’s law Substitute into Gauss’s law: $$$$

Concept & Intuition

Work done by a variable force along a straight line is not simply force times distance — you must integrate. Here, the force depends on position ($F = -K/y^2$), so the work is the area under the $F$‑vs‑$y$ curve between the limits. The negative sign in the force means it opposes the direction of increasing $y$; the integral will naturally account for that.

Step‑by‑Step Solution

1. Recall the definition of work for a one‑dimensional variable force If a force $F(y)$ acts along the $y$-axis, the work done in moving a particle from $y_1$ to $y_2$ is

This is the fundamental relation — no shortcuts when the force isn’t constant.

2. Substitute the given force and limits Here $F(y) = -\dfrac{K}{y^2}$, $y_1 = a$, $y_2 = 2a$. So

3. Evaluate the integral The antiderivative of $y^{-2}$ is $-y^{-1}$ (since $\frac{d}{dy}(-y^{-1}) = y^{-2}$). Thus $$\int_{a}^{2a} y^{-2} \, dy = \left[ -y^{-1} \right]_{a}^{2a} = \left(-\frac{1}{2a}\right) - \left(-\frac{1}{a}\right) = -\frac{1}{2a} + \frac{1}{a} = \frac{1}{2a}. …$$

The key concept here is work as the integral of force with respect to displacement. When a force varies with position, you cannot simply multiply force times distance — you must sum up infinitesimal contributions. The force is given as $F = -\dfrac{K}{y^2}$, acting along the $y$-axis. The negative sign indicates the force opposes the positive $y$-direction (like gravity or an attractive central force). Work done by this force when the particle moves from $y = a$ to $y = 2a$ is:

We integrate carefully, watching signs and limits.

1. Set up the integral The force is $F(y) = -\dfrac{K}{y^2}$. The displacement is along $y$, so work is:

2. Factor out the constant

3. Integrate The antiderivative of $y^{-2}$ is $-y^{-1}$ (since $\frac{d}{dy}(-1/y) = 1/y^2$). So:

Thus:

4. Evaluate the definite integral

Simplify inside:

The key idea here is that magnetic susceptibility ($\chi$) tells us how easily a material gets magnetised in response to an external magnetic field. For a paramagnetic substance, $\chi$ is small and positive. The definition is:

where $M$ is the magnetisation (magnetic moment per unit volume) and $H$ is the applied magnetic field intensity. So we need $M$ first.

1. Find the volume of the cube.

   Side $= 3\,\text{cm} = 3 \times 10^{-2}\,\text{m}$.

   Volume $V = (3 \times 10^{-2})^3 = 27 \times 10^{-6}\,\text{m}^3 = 2.7 \times 10^{-5}\,\text{m}^3$.

2. Magnetisation $M$ is magnetic moment per unit volume:

Simplify:

(Check: $243/2.7 = 90$, then $10^{-6}/10^{-5} = 10^{-1}$, so $90 \times 10^{-1} = 9$.)

3. Applied field intensity $H = 150 \times 10^3\,\text{A/m} = 1.5 \times 10^5\,\text{A/m}$.

4. Susceptibility is then:

   $\chi = \frac{M}{H} = \frac{9}{1.5 \times 10^5} = 6 \times 10^{-5}$ …

The key idea is that electric flux is defined as $\Phi = \vec{E} \cdot \vec{A}$, where $\vec{A}$ is the area vector whose direction is perpendicular to the surface. For a surface lying in the yz-plane, the perpendicular direction is the x-axis. That means only the x-component of $\vec{E}$ contributes to the dot product — the y and z components are parallel to the surface and produce zero flux.

Let’s work through it step by step.

1. Identify the area vector. The surface is on the yz-plane. A vector perpendicular to this plane points along the x-axis. The area is given as $20\ \mathrm{m}^2$, so the area vector is

(We take the positive x-direction by convention; the sign would only matter if we cared about orientation, but here we just want magnitude.)

2. Recall the flux formula. Electric flux through a flat surface is

This is a dot product, so only components of $\vec{E}$ parallel to $\vec{A}$ contribute.

3. Compute the dot product. Given $\vec{E} = 24\hat{i} + 30\hat{j} + 28\hat{k}\ \mathrm{NC^{-1}}$ and $\vec{A} = 20\hat{i}\ \mathrm{m}^2$,

Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = \hat{k} \cdot \hat{i} = 0$, this simplifies to

4. Interpret the result. …

The problem asks for the electric flux through one face of a cube that contains a charged spherical shell at its center. The shell is hollow, with uniform surface charge density $\sigma$, and the cube’s edge is $3r$, so the shell fits entirely inside. The key is to use Gauss’s law and symmetry — no integration over the face is needed.

Why this works: Gauss’s law says the total electric flux through any closed surface equals the net charge enclosed divided by $\epsilon_0$. The cube is a closed surface, and the shell is entirely inside it. Because the shell is centered in the cube, the electric field has the full symmetry of the cube: each face is identical with respect to the charge distribution. Therefore, the total flux through the cube is shared equally among the six faces.

Let’s go step by step.

1. Find the total charge on the spherical shell. The shell has radius $r$ and uniform surface charge density $\sigma$. The surface area of a sphere is $4\pi r^2$, so the total charge is

2. Apply Gauss’s law to the cube. The cube is a closed surface that completely encloses the shell. The net electric flux through the cube is

3. Use symmetry to find the flux through one face. …

The key idea is that electric flux $\Phi = \vec{E} \cdot \vec{A}$, where $\vec{A}$ is a vector perpendicular to the surface with magnitude equal to the area. When a surface is parallel to a coordinate plane, its area vector points along the perpendicular axis. So for a square in the y-z plane, $\vec{A}$ is along $\hat{i}$; for a square in the x-z plane, $\vec{A}$ is along $\hat{j}$. This means only the corresponding component of $\vec{E}$ contributes to the flux.

1. Identify the area vectors.
   • For a square of side $2\ \text{cm} = 0.02\ \text{m}$ parallel to the y-z plane, the area vector is perpendicular to that plane, i.e., along the x-axis.

• For a square parallel to the x-z plane, the area vector is along the y-axis:

2. Compute the fluxes.
   • Flux through the first square:

• Flux through the second square: $$\theta_2 = \vec{E} \cdot \vec{A}_2 = (4\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (4\times10^{-4}\,\hat{j}) = 5 \times 4\times10^{-4} = 20 \times 10^{-4}\ \text{Vm} …$$

Total flux through the closed cube (Gauss's law):

$\phi_{\text{total}} = 4\pi \times 54 \times 10^{3} = 216\pi\times10^{3}~\text{Nm}^2/\text{C}$ …

The key idea is Gauss’s law: the electric flux through a closed surface depends only on the total charge inside, not on the shape of the surface. Here the surface is a cylinder, but that doesn’t matter — we just need the total charge enclosed by the charge distribution itself (which is a sphere of radius $R$). The charge density is spherically symmetric, so we integrate in spherical coordinates.

1. Set up the total charge integral The volume charge density is $\rho(r) = \alpha r$ for $0 \le r \le R$, and zero elsewhere. In spherical coordinates, a volume element is $dV = r^2 \sin\theta \, dr \, d\theta \, d\phi$. The total charge is

2. Separate the integrals The integrand becomes $\alpha r^3 \sin\theta$. Since the limits are independent,

3. Evaluate each factor

   • Radial: $\displaystyle \int_{0}^{R} r^3 \, dr = \frac{R^4}{4}$.
   • Polar: $\displaystyle \int_{0}^{\pi} \sin\theta \, d\theta = [-\cos\theta]_{0}^{\pi} = 2$.
   • Azimuthal: $\displaystyle \int_{0}^{2\pi} d\phi = 2\pi$.

   Multiplying:

   $Q_{\text{enc}} = \alpha \cdot \frac{R^4}{4} \cdot 2 \cdot 2\pi = \alpha \cdot \frac{R^4}{4} \cdot 4\pi = \pi \alpha R^4.$ …

1. Area vector: a square of side $2$ m has area $A=2^2=4\ \text{m}^2$. Being parallel to the x–z plane, its normal is along the y-axis, so $\vec{A}=4\hat{j}\ \text{m}^2$.

2. Field: $\vec{E}=(10\hat{i}+20\hat{j})$ V/m.

3. Flux: …