Electronics (ECE) - MCQ Practice Questions

Understanding:

We need to find the magnetic field intensity $H$ corresponding to an electric field $E_0=10\text{V/m}$ for a plane wave in free space.

Formula:

For a uniform plane wave in free space, the ratio of electric to magnetic field intensity equals the intrinsic impedance:

Step 1: Recall the intrinsic impedance of free space

Step 2: Solve for $H$

Answer:

The magnetic field intensity is approximately $0.0265\text{A/m}$.

Quick Tip:

${\eta }_0=120\pi \approx 377\Omega$ is a standard value to memorise. A common mistake is inverting the ratio and computing $E/H=H$ instead of $H=E/{\eta }_0$.

Understanding:

We must find the electric field magnitude due to a point charge at a given distance.

Formula:

Coulomb's law gives the electric field due to a point charge:

Step 1: Substitute values

Answer:

The magnitude of the electric field at $r=2\text{m}$ is $9\text{V/m}$.

Quick Tip:

$\frac{1}{4\pi {\epsilon }_0}=9\times 10^9$ is a standard constant. The field falls as $1/r^2$, so doubling the distance reduces the field to one-quarter.

Understanding:

We need the magnetic flux density outside an infinitely long current-carrying conductor.

Formula:

Using Ampere's circuital law, the magnetic flux density at distance $r$ from an infinite straight wire is:

Step 1: Substitute values

Answer:

The magnetic flux density at $r=5\text{cm}$ is $40\mu \text{T}$.

Quick Tip:

Note that $\pi$ cancels cleanly in this formula, making the arithmetic simple. A common error is forgetting to convert centimetres to metres before substituting.

Understanding:

We must find the capacitance of a parallel-plate capacitor with a dielectric.

Formula:

The capacitance of a parallel-plate capacitor with a dielectric is:

Step 1: Substitute values

Answer:

The capacitance of the parallel-plate capacitor is approximately $354\text{pF}$.

Quick Tip:

The free-space capacitance (without dielectric) would be $\approx 88.5\text{pF}$; inserting a dielectric of ${\epsilon }_r=4$ multiplies this by exactly $4$.

Understanding:

We need to apply the differential (point) form of Gauss's law to find $\nabla \cdot \mathbf{D}$.

Formula:

The point form of Gauss's law in electrostatics is:

Step 1: Direct application

Step 2: Note the distinction

The relation $\mathbf{D}={\epsilon }_0\mathbf{E}$ in free space gives $\nabla \cdot \mathbf{E}={\rho }_v/{\epsilon }_0$, but the fundamental Maxwell equation involves $\mathbf{D}$, not $\mathbf{E}$, so no ${\epsilon }_0$ appears here.

Answer:

The divergence of $\mathbf{D}$ equals the volume charge density directly.

Quick Tip:

A very common trap is confusing $\nabla \cdot \mathbf{D}={\rho }_v$ with $\nabla \cdot \mathbf{E}={\rho }_v/{\epsilon }_0$. The ${\epsilon }_0$ factor belongs with $\mathbf{E}$, not with $\mathbf{D}$.

Understanding:

We must identify the correct electromagnetic boundary condition for the tangential component of $\mathbf{E}$ at a dielectric-dielectric interface with no free surface charges or currents.

Formula:

From Faraday's law applied at the interface (tangential $\mathbf{E}$ comes from $\oint \mathbf{E}\cdot d\mathbf{l}=0$ for static fields):

Step 1: Derive from integral form

Applying $\oint \mathbf{E}\cdot d\mathbf{l}=0$ around a small rectangular contour straddling the interface and letting the height shrink to zero gives:

Step 2: Distinguish from the normal component condition

The normal component of $\mathbf{D}$ (not $\mathbf{E}$) satisfies the boundary condition:

The relation ${\epsilon }_1{E}_{t1}={\epsilon }_2{E}_{t2}$ applies to normal $D$ components and is a common distractor.

Answer:

The tangential electric field is continuous across any dielectric boundary.

Quick Tip:

Remember: tangential $\mathbf{E}$ is always continuous; normal $\mathbf{D}$ is continuous when there is no free surface charge. The permittivities ${\epsilon }_1,{\epsilon }_2$ appear only in the normal-component condition.

Understanding:

We need the wavelength of a plane wave propagating in free space at a given frequency.

Formula:

The wavelength is related to frequency and wave speed by:

Step 1: Substitute values

Answer:

The wavelength at $3\text{GHz}$ in free space is $0.1\text{m}$, i.e., $10\text{cm}$.

Quick Tip:

A quick rule: at $3\text{GHz}$, $\lambda =10\text{cm}$. Doubling frequency halves the wavelength. This range ($1$–$30\text{GHz}$) is the microwave band.

Understanding:

We must find the intrinsic impedance of a lossless dielectric medium.

Formula:

The intrinsic impedance of a lossless medium is:

Step 1: Substitute values

Answer:

The intrinsic impedance of the medium is approximately $125.7\Omega$.

Quick Tip:

With ${\mu }_r=1$, higher ${\epsilon }_r$ always lowers $\eta$ below ${\eta }_0$. The speed of light in this medium is also reduced by $\sqrt{{\epsilon }_r{\mu }_r}=3$, giving $v=c/3$.

Understanding:

We need to find the $x$-component of $\mathbf{E}$ from the given scalar potential $V$.

Formula:

The electric field is the negative gradient of the potential:

Step 1: Compute partial derivatives

Step 2: Evaluate $E_x$ at $(1,2,3)$

Answer:

The $x$-component of the electric field at $(1,2,3)$ is $-6\text{V/m}$.

Quick Tip:

The negative sign in $\mathbf{E}=-\nabla V$ is frequently dropped by mistake, giving $+6\text{V/m}$. Always remember that $\mathbf{E}$ points in the direction of decreasing potential.