CBSE 12th Standard Physics Subject Electric Charges And Fields Chapter Case Study Questions 2021
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CBSE 12th Standard Physics Subject Electric Charges And Fields Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electrons which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron.
[1.6 x 10-19 C] i.e.
q = ± ne
where n = 1,2,3,4, ....
Hence no body can have a charge represented as \(1.1 e, 2.7 e, \frac{3}{5} e, \text { etc. }\)
Recently, it has been discovered that elementary particles such as protons or neutrons are composed of more elemental units called quarks.
(i) Which of the following properties is not satisfied by an electric charge?(a) Total charge conservation (b) Quantization of charge (c) Two types of charge (d) Circular line of force (ii) Which one of the following charges is possible?
(a) 5.8 x 10-18C (b) 3.2 x 10-18C (c) 4.5 x 10-19C (d) 8.6 x 10-19C (iii) If a charge on a body is 1 nC, then how many electrons are present on the body?
(a) 6.25 x 1027 (b) 1.6 x 1019 (c) 6.25 x 1028 (d) 6.25 x 109 (iv) If a body gives out 109 electrons every second, how much time is required to get a total charge of 1 C from it?
(a) 190.19 years (b) 150.12 years (c) 198.19 years (d) 188.21 years (v) A polythene piece rubbed with wool is found to have a negative charge of3.2 x 1O-7C.Calculate the number of electrons transferred.
(a) 2 x 1012 (b) 3 x 1012 (c) 2 x 1014 (d) 3 x 1014 -
In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled Mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of 1.6 x 10-19 C the charge of the electron. For this, he won the Nobel prize.
(i) If a drop of mass 1.08 x 10-14 kg remains stationary in an electric field of 1.68 x 105 N C-I, then the charge of this drop is(a) 6.40 x 10-19 C (b) 3.2 x 10-19 C (c) 1.6 X 10-19 C (d) 4.8 x 10-19 C (ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are
(a) 4 (b) 3 (c) 5 (d) 8 (iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m-1.If the mass of the drop is 1.6 X 10-3 g, the number of electrons carried by the drop is (g= 10 m s-2)
(a) 1018 (b) 1015 (c) 1012 (d) 109 (iv) The important conclusion given by Millikan's experiment about the charge is
(a) charge is never quantized (b) charge has no definite value (c) charge is quantized (d) charge on oil drop always increases. (v) If in Millikan's oil drop experiment, charges on drops are found to be \(8 \mu \mathrm{C}, 12 \mu \mathrm{C}, 20 \mu \mathrm{C}\) then quanta of charge is
\(\text { (a) } 8 \mu \mathrm{C}\) \(\text { (b) } 20 \mu \mathrm{C}\) \(\text { (c) } 12 \mu \mathrm{C}\) \(\text { (d) } 4 \mu \mathrm{C}\) -
Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in the given figure, the electric field at P is stronger that at Q.
(i) Electric lines of force about a positive point charge are(a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines (ii) Which of the following is false for electric lines of force?
(a) They always start from positive charges and terminate on negative charges (b) They are always perpendicular to the surface of a charged conductor (c) They always form closed loops (d) They are parallel and equally spaced in a region of uniform electric field (iii) Which one of the following pattern of electric line of force in not possible in filed due to stationary charges?
(iv) Electric lines of force are curved
(a) in the field of a single positive or negative charge (b) in the field of two equal and opposite charges (c) in the field of two like charges (d) both (b) and (c). (v) The figure below shows the electric field lines due to two positive charges. The magnitudes EA, EB and Ec of the electric fields at points A, Band C respectively are related as
\(\text { (a) } E_{A}>E_{B}>E_{C}\) \(\text { (b) } E_{B}>E_{A}>E_{C}\) \(\text { (c) } E_{A}=E_{B}>E_{C}\) \(\text { (d) } E_{A}>E_{B}=E_{C}\) -
In practice, we deal with charges much greater in magnitude than the charge on an electron, so we can ignore the quantum nature of charges and imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as continuous charge distribution. There are three types of continuous charge distribution : (i) Line charge distribution (ii) Surface charge distribution (iii) Volume charge distribution as shown in figure.
(I) Statement 1 : Gauss's law can't be used to calculate electric field near an electric dipole.
Statement 2 : Electric dipole don't have symmetrical charge distribution.(a) Statement 1 and statement 2 are true (b) Statement 1 is false but statement 2 is true (c) Statement 1 is true but statement 2 is false (d) Both statements are false (ii) An electric charge of 8.85 X 10-13 C is placed at the centre of a sphere of radius 1 m. The electric flux through the sphere is
(a) 0.2 N C-1 m2 (b) 0.1 N C-1 m2 (c) 0.3 N C-1 m2 (d) 0.01 N C-1 m2 (iii) The electric field within the nucleus is generally observed to be linearly dependent on r. So,
(a) a=O \(\text { (b) } a=\frac{R}{2}\) (c) a=-R \(\text { (d) } a=\frac{2 R}{3}\) (iv) What charge would be required to electrify a sphere of radius 25 cm so as to get a surface charge density of \(\frac{3}{\pi} \mathrm{C} \mathrm{m}^{-2} ?\)
(a) 0.75 C (b) 7.5 C (c) 75 C (d) zero (v) The SI unit of linear charge density is
(a) Cm (b) Cm-1 (c) C m-2 (d) C m-3 -
Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e., \(\sigma=\frac{d q}{d S}\) 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 x 10-22 C m-2 as shown. The intensity of electric field at a point is \(E=\frac{\sigma}{\varepsilon_{0}}\) where \(\varepsilon_{0}\) = permittivity of free space.
(i) E in the outer region of the first plate is\(\text { (a) } 17 \times 10^{-22} \mathrm{~N} / \mathrm{C}\) \(\text { (b) } 1.5 \times 10^{-25} \mathrm{~N} / \mathrm{C}\) \(\text { (c) } 1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) zero (ii) E in the outer region of the second plate is
\(\text { (a) } 17 \times 10^{-22} \mathrm{~N} / \mathrm{C}\) \(\text { (b) } 1.5 \times 10^{-15} \mathrm{~N} / \mathrm{C}\) \(\text { (c) } 1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) zero (iii) E between the plates is
\(\text { (a) } 17 \times 10^{-22} \mathrm{~N} / \mathrm{C}\) \(\text { (b) } 1.5 \times 10^{-15} \mathrm{~N} / \mathrm{C}\) \(\text { (c) } 1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) zero (iv) The ratio of E from right side of B at distances 2 cm and 4 ern, respectively is
(a) 1: 2 (b) 2: 1 (c) 1: 1 \(\text { (d) } 1: \sqrt{2}\) (v) In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is
(a) spherical (b) cylindrical (c) straight line (d) none of these
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CBSE 12th Standard Physics Subject Electric Charges And Fields Case Study Questions 2021 Answer Keys
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(i) (d)
(ii) (b): From, \(q=n e, n=\frac{q}{e}=\frac{3.2 \times 10^{-18}}{1.6 \times 10^{-19}}=20\)
As n is an integer, hence this value of charge is possible.
(iii) (d): Charge on the body is q = ne
\(\therefore\) No. of electrons present on the body is
\(n=\frac{q}{e}=\frac{1 \times 10^{-9} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=6.25 \times 10^{9}\)
(iv) (C): Here, n = 109 electrons per second, Charge given per second,
q = ne = 109 x 1.6 x 10-19C
q = 1.6 X 10-10C
Total charge, Q = 1 C (given)
\(\therefore\) Time required \(=\frac{Q}{q}=\frac{1}{1.6 \times 10^{-10}} \mathrm{~s}=6.25 \times 10^{9} \mathrm{~s}\)
\(\therefore\) \(\frac{6.25 \times 10^{9}}{3600 \times 24 \times 365} \text { year }=198.19 \text { years. }\)
(v) (a): As q = ne, n = \(\frac{3.2 \times 10^{-7}}{1.6 \times 10^{-19}}\)
⇒ n = 2 x 1012electrons. -
(i) (a): As, \(q E=m g \Rightarrow q=\frac{1.08 \times 10^{-14} \times 9.8}{1.68 \times 10^{5}}\)
\(=6.4 \times 10^{-19} \mathrm{C}\)
(ii) (a): \(q=n e \text { or } \Rightarrow n=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\)
(iii) (c) : For the drop to be stationary,
Force on the drop due to electric field = Weight of the drop
qE=mg
\(q=\frac{m g}{E}=\frac{1.6 \times 10^{-6} \times 10}{100}=1.6 \times 10^{-7} \mathrm{C}\)
Number of electrons carried by the drop is
\(n=\frac{q}{e}=\frac{1.6 \times 10^{-7} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=10^{12}\)
(iv) (c)
(v) (d): Millikan's experiment confirmed that the charges are quantized, i.e., charges are small integer multiples of the base value which is charge on electron. The charges on the drops are found to be multiple of 4. Hence, the quanta of charge is 4 \(\mu \)C. -
(i) (a) radially outwards
(ii) (c): Electric lines offorce do not form any closed loops.
(iii) (c) : Electric field lines can't be closed.
(iv) (d) (d) both (b) and (c)
(v) (a) \(\text { (a) } E_{A}>E_{B}>E_{C}\) -
(i) (a): Gauss's law is applicable for any closed surface. Gauss's law is most useful in situation where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over the plane.
Whereas electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance.
So both statements are correct.
(ii) (b): According to Gauss's law, the electric flux through the sphere is
\(\phi=\frac{q_{\mathrm{in}}}{\varepsilon_{0}}=\frac{8.85 \times 10^{-13} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=0.1 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}\)
(iii) (c) : For uniformly volume charge density,
\(E=\frac{\rho r}{3 \varepsilon_{0}}\)
\(E \propto r\)
(iv) (a): r = 25 ern = 0.25 m \(\sigma=\frac{3}{\pi} \mathrm{C} / \mathrm{m}^{2}\)
As, \(\sigma=\frac{q}{4 \pi r^{2}} \Rightarrow q=4 \pi \times(0.25)^{2} \times \frac{3}{\pi}=0.75 \mathrm{C}\)
(v) (b): The line charge density at a point on a line is the charge per unit length of the line at that point
\(\lambda=\frac{d q}{d L}\)
Thus, the SI unit for \(\lambda \text { is } \mathrm{Cm}^{-1} \text {. }\) -
(i) (d):There are two plates A and B having surface charge densities,
\(\sigma_{A}=17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\)
on A and \(\sigma_{B}=-17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\) 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.
\(E_{\mathrm{I}}=E_{A}+E_{B}=\frac{\sigma}{2 \varepsilon_{0}}+\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=0\)
(ii) (d): The electric field in region III is also zero
\(E_{\mathrm{III}}=E_{A}+E_{B}=\frac{\sigma}{2 \varepsilon_{0}}+\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=0\)
(iii) (c) : In region II or between the plates, the electric field
\(E_{\mathrm{II}}=E_{A}-E_{B}=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}\)
\(=\frac{\sigma\left(\sigma_{A} \text { or } \sigma_{B}\right)}{\varepsilon_{0}}=\frac{17.0 \times 10^{-22}}{8.85 \times 10^{-12}}\)
\(E=1.9 \times 10^{-10} \mathrm{~N} \mathrm{C}^{-1}\)
(iv) (c): 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.
(v) (b): In 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.
