GATE EE · Electromagnetic Theory

Electrostatic Fields Practice Questions

Generate GATE-level questions on Electrostatics. Focus on: 1. Coulomb's Law and Electric Field Intensity. 2. Gauss's Law and its applications. 3. Electric Potential, Capacitance, and Energy in electrostatic fields.

35 questions · 15 PYQs · 0 AI practice · GATE EE 2027

Q21GATE 2014MCQ

$C_0$ is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity $\epsilon _r$ , the expression for the modified capacitance is

🚀 Solve in Practice Mode

📖 Explanation

Let A be the area of the parallel plate capacitor and d be the distance between the plates. Let $C_1$ be the capacitance of half portion with air as dielectic medium and $C_2$ be capacitance with a dielectic of permitivity, $\varepsilon _r$ . Then, $C_1=\frac{\varepsilon _0 \left ( \frac{A}{2} \right )}{d}$ (As area become half) and $C_2=\frac{\varepsilon _0 \varepsilon _r \left ( \frac{A}{2} \right )}{d}$ (As area becomes half) Now, these two capacitance will be in parallel if a voltage is applied between the plates as same potential difference will be there between both the capacitance. Equivalent capacitance is

\begin{aligned} C_{eqv} &= C_1+C_2\\ &=\frac{\varepsilon _0 A}{2d}+ \frac{\varepsilon _0 \varepsilon _r A}{2d}\\ &= \frac{\varepsilon _0 A}{2d}(1+ \varepsilon _r)=\frac{C_0}{2}(1+\varepsilon _r) \end{aligned}

Therefore, modified capacitance, $C_{eqv}=\frac{C_0}{2}(1+\varepsilon _r)$

Explanation Figure 2

Explanation Figure 3

Q23GATE 2014MCQ

A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates. If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio $\varepsilon _1:\varepsilon _2$ is

🚀 Solve in Practice Mode

📖 Explanation

Let, A = Area of plates Let $C_1= C_3$ be the capacitance formed with dielectric having dielectric constant $\varepsilon _1$ . $C_{eqv}$ be the equivalent capacitance. $C_2$ be the capacitance formed with dielectic having dielectric constant $\varepsilon _{r_2}$ . Then, $C_1=C_3=\frac{\varepsilon _0\varepsilon _1A}{\frac{d}{4}}=\frac{4\varepsilon _0\varepsilon _1A}{d}$ and $C_2=\frac{\varepsilon _0\varepsilon _2A}{\frac{d}{2}}=\frac{2\varepsilon _0\varepsilon _2A}{d}$ Also, equivalent capacitance $= C_{eqv}$

\begin{aligned} \therefore \;\;\frac{1}{C_{eqv}}&=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{2}{C_1}+\frac{1}{C_2} \\ &(\because \;\;C_1=C_3)(C_1,C_2,C_3 \;\text{all \in series}) \\ \frac{1}{C_{eqv}} &= \frac{2d}{4\varepsilon _0\varepsilon _1d}+\frac{d}{2\varepsilon _0\varepsilon _2d}\\ C_{eqv}&=\frac{2\varepsilon _1\varepsilon _2\varepsilon _0A}{d(\varepsilon _1+\varepsilon _2)} \end{aligned}

Given, $V_{eqv}=$ Total Voltage = 10 VOlt, $V_1=V_3=2$ Volt We know that : $C\propto \frac{1}{V}$

\begin{aligned} &\therefore \; \frac{C_{eqv}}{C_1}=\frac{V_1}{V_{eqv}}=\frac{1}{5} \\ &\frac{2\varepsilon _0\varepsilon _1\varepsilon _2 A}{d(\varepsilon _1+\varepsilon _2)}\times \frac{d}{4\varepsilon _0\varepsilon _1A}=\frac{1}{5} \\ &\frac{\varepsilon _2}{2(\varepsilon _1+\varepsilon _2)}=\frac{1}{5} \\ &5\varepsilon _2=2\varepsilon _1+2\varepsilon _2 \\ &2\varepsilon _1 =3\varepsilon _2\\ &\varepsilon _1:\varepsilon _2=3:2 \end{aligned}