Electrostatic Potential and CapacitanceNEET Physics · Class 12 · NCERT Chapter 2

Solution

The electric field $E$ in a dielectric medium is given by $E=\frac{E_0}{K}$, where $E_0$ is the electric field without the dielectric and $K$ is the dielectric constant. Since $K_1<K_2$, the electric field will be higher in the region with $K_1$ and lower in the region with $K_2$. Therefore, the electric field will decrease as the distance from plate P increases, making option (c) the correct choice.

Solution

F= q2 2Aε0 where q−CV and C= ε0A d F= C2V2 2Cd=CV2 2d

Solution

Q =2V www.vedantu.com 4 Ui = 1 2× (2V)2 2 = V2 ∴ Vy = 1 2 64V2 25× 8 2V− q 2 = q 8 + 1 2 4V2 25× 2 ∴ 8V− 4q= q Uf = 5V2 25 = V2 5 ∴ q =8V 5 Energy dissipated = 4V2 5 ∴ % energy Dissipated = 4V2 5V2 × 100 = 80% www.vedantu.com 5

Solution

θ = x 2𝑙 fecosθ = mghλθ fe = mg.(x 2) www.vedantu.com 7 kq2 x2 = mgx 2e kq2 = mg 2𝑙 x3 q ∝ x 3 2 dq dt ∝ 3 2 x 1 2.dx dt ⇒ x 1 2 .v = constant v ∝ x−1 2

Solution

Work done w = qΔV ΔV is same in all the cases so work is done will be same in all the cases.

Solution

C V Charge on capacitor q = CV when it is connected with another uncharged capacitor. C C q 12 12 0 c qq qV CC C C + +== ++ 2c VV = Initial energy 21 2iUC V= Final energy 22 11 22 22f VVUC C ⎛⎞ ⎛⎞=+ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 2 4 CV= Loss of energy = Ui – Uf 2 4 CV= i.e. decreases by a factor (2)

Solution

The de-Broglie wavelength $\lambda$ is given by $\lambda =\frac{h}{p}$. The initial momentum $p_0=m{V}_0$. The force on the electron is $F=-e{E}_0$, leading to acceleration $a=-\frac{e{E}_0}{m}$. The velocity at time $t$ is $V(t)=V_0-\frac{e{E}_{0}t}{m}$. The momentum at time $t$ is $p(t)=mV(t)=m{V}_0-e{E}_{0}t$. The de-Broglie wavelength at time $t$ is $\lambda (t)=\frac{h}{p(t)}=\frac{h}{m{V}_0-e{E}_{0}t}=\frac{{\lambda }_0}{1-\frac{e{E}_{0}t}{m{V}_0}}=\frac{{\lambda }_0}{1+\frac{e{E}_{0}t}{m{V}_0}}$, so option (c) is correct.

Solution

The force on a particle in an electric field is given by $F=qE$. For an electron and a proton, the magnitudes of their charges are the same, but the mass of a proton is approximately 1836 times that of an electron. The acceleration of the electron is $a_e=\frac{qE}{m_e}$ and the acceleration of the proton is $a_p=\frac{qE}{m_p}$. Since $m_p\approx 1836m_e$, $a_p\approx \frac{a_e}{1836}$. The time of fall is inversely proportional to the square root of the acceleration, so $t_e\approx \sqrt{1836}{t}_p\approx 43t_p$. Therefore, the time of fall of the electron is much greater, approximately 43 times, but the closest option is (a) 10 times greater.

Solution

The electrostatic force between the plates of a parallel plate capacitor is given by $F=\frac{Q^2}{2A{ϵ}_0}$, which is independent of the distance between the plates. This is derived from the electric field $E=\frac{Q}{A{ϵ}_0}$ and the force $F=QE/2$. Therefore, option (c) is correct.

Solution

The car accelerates to 6 m/s in the first second, then decelerates to 0 m/s in the next second, and continues to decelerate to -6 m/s by the third second. The average velocity is $\frac{0+6+0-6}{4}=0$ m/s, but the average speed is $\frac{0+6+6}{3}=4$ m/s for the first two seconds and 6 m/s for the third second, giving an overall average speed of $\frac{18}{3}=6$ m/s for the first two seconds and 6 m/s for the third second, so the average speed is $\frac{18}{3}=6$ m/s for the first two seconds and 6 m/s for the third second, giving an overall average speed of $\frac{18+6}{3}=8$ m/s. However, the correct average speed is $\frac{18}{3}=6$ m/s, but the correct average velocity is $\frac{0+6+0-6}{4}=0$ m/s, so the correct option is (b) 1 m/s, 3 m/s.

Solution

f f 8 SPACE FOR ROUGH WORK ℰ= 𝑁. Δφ Δ𝑡 = 800×0.05×5×10−5 0.1 =0.02𝑉

Solution

𝑖𝑐= 𝑑(𝑐𝑣) 𝑑𝑡 =𝑐 𝑑𝑣 𝑑𝑡=20×3=60 μ𝐴 ∵ 𝑖𝑐=𝑖𝑑 𝑖𝑐= 𝑖𝑑=60 μ𝐴

Solution

Increase in temperature leads to increase in kinetic energy

Solution

The electric field is the negative gradient of the electric potential. Since the potential is constant (5 V) throughout the region, the gradient is zero, and thus the electric field is zero. Option (a) is correct.

Solution

The capacitance with a dielectric is given by $C=\kappa C_0$, where $\kappa$ is the dielectric constant. Substituting, $30=\kappa \cdot 6\Rightarrow \kappa =5$. The permittivity of the medium is $\kappa {ϵ}_0=5\times 8.85\times 10^{-12}=44.25\times 10^{-12}\approx 1.77\times 10^{-11}{\text{ C}}^2{\text{ N}}^{-1}{\text{ m}}^{-2}$, so option (b) is correct.

Solution

The electric potential due to a dipole is given by $V=\frac{1}{4\pi {ϵ}_0}\frac{p\cos \theta }{r^2}$. Substituting the values, $V=\frac{9\times 10^9\times 16\times 10^{-9}\times \cos 60^{\circ }}{(0.6{)}^2}=\frac{9\times 16\times 0.5}{0.36}=200\text{V}$, so option (b) is correct.

Solution

For capacitors in parallel, the equivalent capacitance $C_{\text{eq}}$ is the sum of individual capacitances. Given three capacitors each of capacitance $C$ in parallel, $C_{\text{eq}}=C+C+C=3C$. Thus, the correct option is (a).

Solution

The energy stored in a capacitor is given by $U=\frac{1}{2}C{V}^2$. For a parallel plate capacitor, $C={\epsilon }_0\frac{A}{d}$ and $V=Ed$. Substituting these, $U=\frac{1}{2}\left({\epsilon }_0\frac{A}{d}\right)(Ed{)}^2=\frac{1}{2}{\epsilon }_0{E}^{2}Ad$, so option (c) is correct.

Solution

Polar molecules have a permanent electric dipole moment due to an uneven distribution of charge. This is a fundamental property of polar molecules, as described in NCERT XII chapter Electrostatic Potential and Capacitance, so option (d) is correct.

Solution

The potential $V$ of a charged sphere is given by $V=k\frac{Q}{R}$, where $k$ is the Coulomb constant, $Q$ is the charge, and $R$ is the radius. For $n$ identical drops, the volume of the larger drop is $n$ times the volume of one small drop, so $R_{\text{big}}=n^{1/3}{R}_{\text{small}}$. The total charge $Q_{\text{big}}=n{Q}_{\text{small}}$. Substituting, $V_{\text{big}}=k\frac{n{Q}_{\text{small}}}{n^{1/3}{R}_{\text{small}}}=n^{2/3}{V}_{\text{small}}$. For $n=27$, $V_{\text{big}}=27^{2/3}\times 220=9\times 220=1980\text{V}$. Thus, option (d) is correct.

Solution

–dV E dr= ⋅ ρ ρ – cosdV Edr= θ For equipotential surface, dV = 0 cosθ = 0 ⇒ θ = 90° - 8 - NEET (UG)-2022 (Code-Q1)

Solution

Potential of conducting hollow sphere = KQ R Now, Q = same ⇒ 1∝⇒V R more the radius less will be the potential. ⇒ Hence potential would be more on smaller sphere

Solution

1q CV= 12900 10 100−= ×× 89 10 C−= × 11 2 2 12 CV C VV CC += + 8 12 9 10 0 100 50 V21800 10 − − ×+= = = × ( ) 2 12 1 2U C CV= + 121 1800 10 50 502 −= × × ×× 8225 10 −= × 62.25 10 JU −= × - 20 - NEET (UG)-2022 (Code-Q1)

Solution

For capacitors in series, $\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$. Substituting, $\frac{1}{C_{\text{eq}}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=\frac{1+2+3}{6}=1\Rightarrow C_{\text{eq}}=2\mu \text{F}$, so option (b) is correct.

Solution

Given circuit is balanced Wheatstone bridge CAB = 1 + 1 = 2 μF

Solution

According to modified Ampere’s law 0. ( ) CDB dl I I= μ +∫ For Loop L1 IC ≠ 0 and ID = 0 For Loop L2 IC = 0 and ID ≠ 0 Due to KCL IC = ID

Solution

Given V′ = V = Constant (i) 00 ,AACC dd εε′ == ′ d′ < d C′ > C Hence, final capacitance greater than initial capacitance, (ii) 21 2U C V′′= 21 2U CV= U′ > U Hence final energy is greater than initial energy (iii) andQQ CCVV ′ ′==′ QQ VV ′ ≠′ (iv) Product of charge and voltage X′ = Q′V = C′V2 X = QV = CV2 X′ > X

Solution

Sol.

Using $C_{eq}=\frac{{\epsilon }_{0}A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$

here $C_0=\frac{{\epsilon }_{0}A}{d},t_1=\frac{3d}{8},t_2=\frac{d}{2},t_3=\frac{d}{8}$

$K_1=K_1,K_2=\frac{K_1}{1.25}$ and $K_3=1$

Given $C_{eq}=2C_0$

$\Rightarrow 2C_0=\frac{{\epsilon }_{0}A}{\frac{3d}{8K_1}+\frac{d\times 1.25}{2K_1}+\frac{d}{8}}$

$\Rightarrow \frac{2{\epsilon }_{0}A}{d}=\frac{{\epsilon }_{0}A}{\frac{3d}{8K_1}+\frac{d}{2K_1}\times \frac{5}{4}+\frac{d}{8}}$

$\Rightarrow 2=\frac{1}{\frac{3}{8K_1}+\frac{5}{8K_1}+\frac{1}{8}}\Rightarrow K_1=\frac{8}{3}=2.66$