Q141JEE Main 2019MCQ

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d ≪ a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is : [9-Jan-2019 Shift 1]

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📖 Explanation

$y/x$ = $d/a$ y = $d/a$ x dy = $d/a (dx)$ $1/{dc}$ = $y/{KE.adx}$ + $(d-y)/{∊_0adx}$ $1/{dc}$ = $1/{∊_0 adx}$ $(y/k + d - y)$ ∫ dc = ∫ $\frac{∊_0adx}{y/k + d-y}$ c = $∊_0a$ . $a/2 ∫^{d}_{0}$ $\frac{dy}{d+y(1/k-1)}$ = $\frac{∊_0A=a=a^2}{(1/k-1)d}$ $[ln (d+y(1/k-1))]^0_0$ = $\frac{k∊_0a^2}{(1-k)d}$ ln $({d+d(1/k-1)}/d)$ = $\frac{k∊_0a^2}{(1-k)d}$ ln $(1/k)$ = $\frac{k∊_0a^2lnk}{(k-1)d}$

Q143JEE Main 2019MCQ

The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is:

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📖 Explanation

$Q=C V(n+1)$ $\ \frac{q}{K C}=\ \frac{Q-q}{n C}$ $q\ (\ \frac{n+K}{n K}\ )=\ \frac{Q}{n}$ $q=\ \frac{Q K}{n+K}$ $V^{,}=\ \frac{q}{K C}$ $= \frac{1}{C} \ \frac{Q}{n+K}=\ \frac{V(n+1)}{(n+K)}$

Q144JEE Main 2018MCQ

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant $K=5/3$ is inserted between the plates, the magnitude of the included charge will be :[Main 8 April 2018]

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📖 Explanation

$q'=q(1-1/k)$ $=kcv(1-1/k)$ $=5/3\times 90\times 20(1-3/5)=1.2nc$

Q145JEE Main 2018MCQ

A capacitor $C _1=1.0 \mu F$ is charged up to a voltage $V =60 V$ by connecting it to battery B through switch ( 1). Now $C _1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C _2=3.0 \mu F$ and $C _3=6.0 \mu F$ through switch ( 2), as shown in the figure. The sum of final charges on $C _2$ and $C _3$ is :

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Q146JEE Main 2018MCQ

The equivalent capacitance between A and B in the circuit given below, is[Main 15 April 2018 S1]

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Q147JEE Main 2018MCQ

In the following circuit, the switch S is closed at $t=0$ . The charge on the capacitor $C_1$ as a function of time will be given by $(C_{eq}=\frac{C_1C_2}{C_1+C_2})$ [Main 16 April 2018 S1]

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Q148JEE Main 2018MCQ

A parallel plate capacitor with area $200 cm^2$ and separation between the plates 1.5 cm , is connected across a battery of emf V . If the force of attraction between the plates is $25 \times 10^{-6} N$ , the value of V is approximately : $(E_0=8.5 \times 10^{-12} \;\frac{C^2}{N \cdot m^2})$

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Q149JEE Main 2017MCQ

In the given circuit diagram when the current reaches steady state in the circuit,the charge on the capacitor of capacitance C will be :[Main 2017]

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📖 Explanation

It steady state, current through AB = 0 $\Rightarrow V_{AB}=V_{CD}$ $\Rightarrow V_{AB}=(\epsilon /{r+r_2}\times r_2)-V_{CD}$ $\Rightarrow Q_c=CV_{AB}$ $=CE(r_2/{r+r_2})$

Q150JEE Main 2017MCQ

A combination of parallel plate capacitors is maintained at a certain potentialdifference.When a 3 mm thick slab is introducedbetween all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.[Main 9 April 2017]

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Q151JEE Main 2017MCQ

The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 µC charge, its radius will be: $[Take: $1/{4π∊_0}=9×10^9N-m^2\text{/}C^2$ ]$[Main 8 April 2017]

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📖 Explanation

Energy of sphere $=Q^2/{2C}$ $4.5=\frac{16\times 10^{-12}}{2C}$ $C={16\times 10^{-12}}/9=4\pi \epsilon _0R$ $R={16\times 10^{-12}}/9=1/{4\pi \epsilon _0}$ $=9\times 10^9\times 16/9\times 10^{-12}$ $=16\times 10^{-3}=16$ mm

Q152JEE Main 2017MCQ

A capacitance of 2µF is required in an electrical circuit across a potential difference of 1.0 kV.A large number of 1µF capacitors are available which can withstand a potential difference of not more than 300 V.The minimum number of capacitors required to achieve this is :[Main 2017]

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📖 Explanation

To hold 1 KV potential difference minimum four capacitors are required in series $\Rightarrow C_1=1/4$ for one series So for Ceq to be 2μF, 8 parallel combinations are required.⇒ Minimum no. of capacitors = 8 × 4 = 32

Q153JEE Main 2016MCQ

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be $1\mu F$ [Main 10 Apr 2016]

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📖 Explanation

${8\times 12}/18=4\mu F$ $4\mu F+4\mu F=8\mu F$ $\frac{8\times 1}{8+1}=8/9\mu F$ $\frac{8\times 4}{8+4}=32/12=8/3\mu F$ $8/3+8/9={24+8}/9=32/9\mu C$ $\frac{32/9\times C}{32/9+C}=1\Rightarrow 32/9\times C=32/9+C$ $\Rightarrow C({32-9}/9)=32/9$ $C=32/23\mu F$

Q154JEE Main 2016MCQ

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal : [Main 2016]

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📖 Explanation

Charge on 9µF capacitor = 18µCCharge on 4µF capacitor = 24µC $\therefore Q=24+18=42\mu C$ $\therefore {KQ}/r^2={9\times 10^9\times 42\times 10^{-6}}/(30)^2=420$ N/C

Q155JEE Main 2016MCQ

Three capacitors each of 4 µF are to be connected in such a way that the effective capacitance is 6 µF. This can be done by connecting them :[Main 9 Apr 2016]

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📖 Explanation

where $c=4\mu F$

Q156JEE Main 2015MCQ

In the given circuit, charge $Q_2$ on the 2μF capacitor changes as C is varied from 1μF to 3μF. $Q_2$ as a function of 'C' is given properly by : (figures are drawn schematically and are not to scale)[Main 2015]

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📖 Explanation

Let the potential at P be V,Then,C(E-V)=1 × V + 2 × V(we taken C inμF)Or, $V=\frac{CE}{3+C}$ $\therefore Q_2=\frac{2CE}{3+C}$

Q157JEE Main 2014MCQ

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric constant 2.2 between them.When the electric filed in the dielectric is $3\times 10^4$ V/m,the charge density of the positive plate will be close to:[Main 2014]

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📖 Explanation

By formula of electric filed between the plate of a capacitor $E= \sigma /K\epsilon _0$ $\Rightarrow \sigma =EK\epsilon _0=3\times 10^4\times 2.2\times 8.85\times 10^{-12}$ $=6.6\times 8.85\times 10^{-12}$ $=5.841\times 10^{-7}$ $≅6.10^{-7} C/m^2$

Q158JEE Main 2013MCQ

Two capacitors $C_1$ and $C_2$ are charged to 120 V and 200 V respectively. It is found that by connectingthem together the potential on each one can be made zero. Then :[Main 2013]

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📖 Explanation

$120C_1=200C_1$ $6C_1=10C_2$ $3C_1=5C_2$

Q159JEE Main 2012MCQ

The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $\tau$ of this circuit lies between

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📖 Explanation

During discharging of a capacitor, ${\text{V}}= {\text{V}}_0 e^{-\;\frac{t}{\tau }}$ At ${\text{t}}=\tau$ , ${\text{V}}=\;\frac{V_0}{e}=0.37 {\text{V}}_0$ From the graph, ${\text{t}}=0, {\text{V}}_0=25 {\text{V}}$ $\therefore {\text{V}}=0.37 \times 25=9.25 {\text{V}}$ This voltage will occur at time between $100 {\ sec}$ and $150 {\ sec}$ . Hence, time constant $\tau$ of this circuit lies between $100 {\ sec}$ and $150 {\ sec}$ .

Q160JEE Main 2010MCQ

Let $C$ be the capacitance of a capacitor discharging through a resistor $R$ . Suppose $t_1$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_2$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_1 / t_2$ will be

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📖 Explanation

Initial energy of capacitor, $E_1=\;\frac{q_1^2}{2 C}$ Final energy of capacitor, $E_2=\;\frac{1}{2} E_1=\;\frac{q_1^2}{4 C}= (\;{\;{q_1}/{{\sqrt{2}}}}/{2 C}) ^2$ $\therefore t_1=$ time for the charge to reduce to $\;{1}/{{\sqrt{2}}}$ of its initial valueand $t_2=$ time for the charge to reduce to $\;\frac{1}{4}$ of its initial valueWe have, $q_2=q_1 e^{-t / C R}$ $\Rightarrow \ln (\;\frac{q_2}{q_1}) =-\;\frac{t}{C R}$ $\therefore \ln (\;{1}/{{\sqrt{2}}}) =\;\frac{-t_1}{C R} . . .(1)$ and $\ln (\;\frac{1}{4}) =\;\frac{-t_2}{C R} . . .(2)$ By (1) and $(2), \;\frac{t_1}{t_2}=\;{\ln ({1}/{{\sqrt{2}}}) }/{\ln (\;\frac{1}{4}) }$ $=\;\frac{1}{2} \;\frac{\ln (\frac{1}{2}) }{2 \ln (\;\frac{1}{2}) }=\;\frac{1}{4}$