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Electric Charges and Fields — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields4 Marks
Question
Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e., $\sigma=\frac{\text{dq}}{\text{dS}}.$ Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of $17.0 \times 10^{-22}Cm^{-2 }$ as shown. The intensity of electric field at a point is $\text{E}=\frac{\sigma}{\epsilon_0},$ where $\epsilon_0=$ permittivity of free space.
  1. $E$ in the outer region of the first plate is:
  1. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-25} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. $E$ in the outer region of the second plate is:
  1. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-15} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. $E$ between the plates is:
  1. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-15} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. The ratio of $E$ from right side of $B$ at distances $2\ cm$ and $4\ cm,$ respectively is:
  1. $1 : 2$
  2. $2 : 1$
  3. $1 : 1$
  4. $1:\sqrt{2}$
  1. ln order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is:
  1. Spherical.
  2. Spherical.
  3. Straight line.
  4. None of these.
✓

Answer

  1. $(d)$ Zero.
There are two plates $A$ and $B$ having surface charge densities,

$\sigma_\text{A}=17.0\times10^{-22}\text{C/m}^2$
on $A$ and $\sigma_\text{B}=-17.0\times10^{-22}\text{C/m}^2$on $B,$ respectively. According to Gauss' theorem, if the plates have same surface charge density but having opposite signs, then the electric field in region $I$ is zero.
$\text{E}_\text{I}=\text{E}_\text{A}+\text{E}_\text{B}$
$=\frac{\sigma}{2\epsilon_0}+\Big(-\frac{\sigma}{2\epsilon_0}\Big)=0$
  1. $(d)$ Zero.
The electric field in region $III$ is also zero.
$\text{E}_\text{III}=\text{E}_\text{A}+\text{E}_\text{B}$
$=\frac{\sigma}{2\epsilon_0}+\Big(-\frac{\sigma}{2\epsilon_0}\Big)=0$
  1. $(c) 1.9 \times 10^{-10} N/C$
​​​​​​​​​​​​​​In region $II$ or between the plates, the electric field.
$\text{E}_\text{II}=\text{E}_\text{A}-\text{E}_\text{B}$
$=\frac{\sigma}{2\epsilon_0}+=\frac{\sigma}{2\epsilon_0}$
$=\frac{\sigma(\sigma_\text{A}\text{ or }\sigma_\text{B})}{\epsilon_0}=\frac{17.0\times10^{-22}}{8.85\times10^{-12}}$
$E = 1.9 \times 10^{-10} N/C$​​​​​​​
  1. $(c) 1 : 1$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Since electric field due to an infinite$-$plane sheet of charge does not depend on the distance of observation point from the plane sheet of charge. So, for the given distances, the ratio of $E$ will be $1 : 1.$​​​​​​​
  1. $(b)$ Spherical.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​ln order to estimate the electric field due to a thin finite plane metal plate, we take a cylindrical cross$-$sectional area $A$ and length $2r$ as the gaussian surface.

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If double slit apparatus is immersed in a liquid of refractive index, $\mu$ the wavelength of light reduces to $\lambda$ and fringe width also reduces to $\beta=\frac{\beta}{\mu}$.
The given figure shows a double $-$ slit experiment in which coherent monochromatic light of wavelength $\lambda$ from a distant source is incident upon the two slits, each of width $\text{w}(\text{w}>>\lambda)$ and the interference pattern is viewed on a distant screen. A thin piece of glass of thickness $ t$ and refractive index $n$ is placed between one of the slit and the screen, perpendicular to the light path.
  1. ln Young's double slit interference pattern, the fringe width.
  1. Can be changed only by changing the wavelength of incident light.
  2. Can be changed only by changing the separation between the two slits.
  3. Can be changed either bychangingthe wavelength or by changing the separation between two sources.
  4. Is a universal constant and hence cannot be changed.
  1. If the width w ofone of the slits is increased to $2w,$ the become the amplitude due to slit.
  1. $1.5\text{a}$
  2. $\frac{\text{a}}{2}$
  3. $2\text{a}$
  4. No change.
  1. ln $\text{YDSE},$ let $A$ and $B$ be two slits. Films of thicknesses $t_A$ and $t_B$ and refractive indices $m_A$ and $m_B$ are placed in front of $A$ and $B,$ respectively. If $\mu_\text{A}\text{t}_\text{A}=\mu_\text{B}\text{t}_\text{B}$ then the central maxima will:
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  2. Shift towards $A$.
  3. Shift towards $B.$
  4. Shift towards $A$ if $t_B = t_A$ and shift towards $B$ if $t_B < t_A$​​​​​​​
  1. ln Young's double slit experiment, a third slit is made in between the double slits. Then:
  1. Fringes of unequal width are formed.
  2. Contrast between bright and dark fringes is reduced.
  3. Intensity of fringes totally disappears.
  4. Only bright tight is observed on the screen.
  1. ln Young's double slit experiment, if one of the slits is covered with a microscope cover slip, then:
  1. Fringe pattern disappears.
  2. The screen just gets illuminated.
  3. In the fringe pattern, the brightness of the bright fringes will decreases and the dark fringes will become more dark.
  4. Bright fringes will be more bright and dark fringes will become more dark.
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Distance between two successive bright or dark fringes is called fringe width.
$\beta=\text{Y}_\text{n+1}-\text{Y}_\text{n}=\frac{(\text{n}+1)\lambda\text{D}}{\text{d}}-\frac{\text{n}\lambda\text{D}}{\text{d}}=\frac{\lambda\text{D}}{\text{d}}$
Fringe width is independent of the order of the maxima. If whole apparatus is immersed in liquid of refractive index $\mu$ then $\beta=\frac{\lambda\text{D}}{\mu\text{d}} \ ($fringe width decreases$)$. Angular fringe width $(\theta)$ is the angular separation between two consecutive maxima or minima
$\theta=\frac{\beta}{\text{D}}=\frac{\lambda}{\text{d}}$
ln the arrangement shown in figure, slit $S_3$ and $S_4 $ are having a variable separation $Z$. Point $O$ on the screen is at the common perpendicular bisector of $S_1,S_2$ and $S_3,S_4.$
  1. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double $-$ slit experiment, is:
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  4. Disappears.
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  8. Become narrower.
  9. Disappear.
  10. When the complete Young's double slit experiment is immersed in water, the fringes.
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  12. Halved
  13. Doubled
  14. Quadrupled
  15. ln Young's double slit experiment, if the separation between the slits is halved and the distance between the slits and the screen is doubled, then the fringe width compared to the unchanged one will be.
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  19. Darker
  20. ln Young's double $- $ slit experiment if yellow light is replaced by blue light, the interference fringes become.
  21. Infinite
  22. Five
  23. Three
  24. Zero
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 $0.51$ $0.39$
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  2. $1 \times 10^{-5}kg$
  3. $3 \times 10^{-5}kg$
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