Physics MCQs for NEET — Practice Questions with Answers

$\frac{1}{C}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

⇒ $C=\frac{2}{3}F$ 

By using, common potential $V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

⇒ $20=\frac{2\times 200+C_2\times 0}{2+C_2}$ ⇒ C₂ = 18 μF

When two uncharged identical capacitors are connected in parallel, the equivalent capacitance becomes twice the individual capacitance. Since the total charge remains the same, the energy stored in the combined system decreases by a factor of 2.

When a parallel plate air capacitor connected to a coil of emf V, then charge stored will be

q=CV

=> V=q/C

Also, energy stored is U=$\frac{1}{2}$CV²=$\frac{{\displaystyle q^2}}{{\displaystyle 2C}}$

As the battery is disconnected from the capacitor the charge will not be destroyed i.e. q'=q with the introduction of dielectric in the gap of the capacitor the new capacitance will be
C'=CK

=> V'=q/C'=q/CK

The new energy stored  will be

U'=$\frac{{\displaystyle q^2}}{{\displaystyle 2CK}}$

ΔU=U'-U=$\frac{{\displaystyle q^2}}{{\displaystyle 2c}}$$\left(\frac{1}{K-1}\right)$

= $\frac{1}{2}$CV^{2$\left(\frac{1}{K-1}\right)$}

So, option (a),(b),(c) is correct but (d) is incorrect

$$\begin{gathered} Given, \\ Potential in a region; V=6xy-y+2yz \\ \overrightarrow{E}=-\frac{\delta V}{\delta x}\hat{i}-\frac{\delta V}{\delta y}\hat{j}-\frac{\delta V}{\delta z}\hat{k} \\ \overrightarrow{E}=-\frac{\delta }{\delta x}\left(6xy-y+2yz\right)\hat{i}-\frac{\delta }{\delta y}\left(6xy-y+2yz\right)\hat{j}-\frac{\delta }{\delta z}\left(6xy-y+2yz\right)\hat{k} \\ \overrightarrow{E}=-\left(6y\right)\hat{i}-\left(6x+2z\right)\hat{j}-\left(2y\right)\hat{k} \\ At \left(1, 1, 0\right), \overrightarrow{E}=-\left(6\right)\hat{i}-\left(6\right)\hat{j}-\left(2\right)\hat{k} \end{gathered}$$

Force between plates of parallel capacitor 

F=qE=q[$\sigma$/2${\epsilon }_0$]

∴Surface charge density $\sigma$=q/A

∴F=q[q/2A$\epsilon$_o]

=>F=q²/2A$\epsilon$_o

So, net charge across a capacitor, q=CV

F=^{$\frac{C^2{V}^2}{2A{\epsilon }_0}$}      <sub>$\left[C=\frac{A{\epsilon }_0}{d}\right]$

=>$F=\left(\frac{A{\epsilon }_0}{d}\right)$  x $\frac{{\displaystyle C{V}^2}}{{\displaystyle 2A{\epsilon }_0}}$   

^{$=\frac{{\displaystyle C{V}^2}}{{\displaystyle 2d}}$}</sub>

For a uniformly charged spherical conductor, the electric potential inside the sphere is constant and equal to Q/4πϵ₀R, while the electric field inside is zero due to the cancellation of fields from different parts of the charge distribution.

We know 

F=qE ...(i)

E=-dV/dr

E_x=$\delta$V/$\delta$x=6-8y

E_y=$\delta$V/$\delta$y=-8x-8+6z ...(ii)

E_z=6y

Above values of E_x,Ey and E_z at (1,1,1) are 

E_{x=$6-8 x \left(1\right)=-2$}

Ey=-8(1)-8+6(1)=-10 

E_{z=$6 x1=6$}

So, E_{net=√(-2)2+(10)2+(6)2}

_{=√4+100+36=√140 =>√35x4=2√35 N/C}

_{So, F=qEnet=2(2√35)=4√35N}

If potential at centre is zero, then

$V_1+V_2+V_3+V_4=0$

$-\frac{kQ}{r}-\frac{kq}{r}+\frac{k2Q}{r}+\frac{k2q}{r}=0$

$-Q-q+2q+2Q=0$

$Q=-q$

Charge flows from high potential to low

potential

              $$\begin{gathered} \frac{{\mathrm{KQ}}_1}{3}=\frac{{\mathrm{KQ}}_2}{1} \\ {\mathrm{Q}}_1=3{\mathrm{Q}}_2 \end{gathered}$$

Also,       $$\begin{gathered} {\mathrm{Q}}_1+{\mathrm{Q}}_2=4\times {10}^{-2} \\ 4{\mathrm{Q}}_2=4\times {10}^{-2} \end{gathered}$$

                   ${\mathrm{Q}}_2={10}^{-2}$

and              ${\mathrm{Q}}_1=3\times {10}^{-2}\mathrm{C}$