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Answer: Question 54. Students can solve NCERT Class 12 Physics Electric Charges and Fields MCQs Pdf with Answers to know their […] Answer: At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πrl, where l is the length of the cylinder. Electric flux. Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. Hence obtain the expression for the energy density of the electric field. Semiconductors are acted as the fundamental core materials. (a) \(\overrightarrow { \tau } \) is perpendicular to \(\overrightarrow { p } \) (a) (i) Energy of a parallel plate capacitor. By symmetry, electric field E points outwards normal to the sheet. Answer: Question 48. (Comptt. (i) For first case, θ = 0, cos 0 = 1 Torque, acting on the dipole is, τ = pE sin θ. (ii) Angle of square plane with x-axis = 30° Question 35. A small metal sphere carrying charge +Q is located at the centre of a spherical cavity in a large uncharged metallic spherical shell. Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. Electric dipole moment: It is the product of the magnitude of either charge and distance between them. (ii) Torque acting on these cases Answer: Answer: or Strength of electric dipole is called dipole moment. Obtain an expression for the electric intensity E at a point on the axis of the ring. Derive an expression for the electric field due to an infinitely long straight uniformly charged wire. All India 2015) (ii) the orientation of the dipole for which the torque acting on it becomes maximum. Electric field due to a uniformly charged spherical shell: (Delhi 2013). Using Gauss’s law, derive an expression for an electric field at a point outside the shell. Answer: Zero because the net charge of an electric dipole (+ q and – q) is zero. This wire is enclosed by a Gaussian hollow surface. Electrostatic field inside a conductor should be zero because of the absence of charge. Answer: opposite to \(\overrightarrow{\mathbf{E}}\) . Question 56. Electric Charges and Fields Important Questions for CBSE Class 12 Physics Coulombs Law, Electrostatic Field and Electric Dipole 1.Electric Charge Charge is the property associated with matter due to which it produces and experiences electric and magnetic effect. Is the potential difference VA – VB positive, negative or zero? Answer: In this lesson, we'll define perfusion. Interpret the graphs obtained. Answer: Answer: Point lies inside both the spherical shells. (All India 2013) At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since at every point, and its magnitude is constant, since it depends only on r. The surface area of the cylinder. Answer: Electric flux ϕ by qenclosed Question 61. A = 10 × 10 × 10-4m2, (a) An electric dipole of dipole moment \(\overrightarrow { p } \) consists of point charges + q and – q separated by a distance 2a apart. Given a uniform electric field \(\overrightarrow{\mathrm{E}}=4 \times 10^{3} \hat{i}\) N/C. Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface As point P2 lies inside the metal, therefore electric field at point P2 is zero. Write its S.I. (Delhi 2012) Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density X. …. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Question 34. They occur only in polar molecules such as HCl. A sphere S1 of radius r1 encloses a net charge Q. It is a scalar quantity. (a) Derive the expression for the energy stored in a parallel plate capacitor. (a) stable, Question 2. Let at any instant when charge on capacitor be q, the potential difference between its plates be, Since is uniform, hence the dipole does not undergo any translatory motion. Question 12. (a) Define electric flux. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. Draw the electric field lines between the surface and the charge. [G16 Rev. CBSE Class 12 Physics Notes Chapter 14 Semiconductor Electronic Material Devices and Simple Circuits In order to define the Semiconductors, these are used in solid-state electronic devices such as transistors, diodes, etc. (b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. Figure \(\PageIndex{5}\): Molecules That Possess a Dipole Moment Partially Align Themselves with an Applied Electric Field. Let a point P be outside the shell with radius vector. Question 14. (i) Field Outside Shell : Hence, E(x) = 0 \(\overrightarrow{\mathrm{P}} \| \overrightarrow{\mathrm{E}}\). Let σ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find \(\overrightarrow{\mathrm{E}}\) We choose cylindrical Gaussian surface of cross-sectional area A and length 2r with its axis perpendicular to the sheet. (b) Draw a graph of E versus r for r >> a. What would be the flux through the same square if the plane makes a 30° angle with the x-axis? Answer: (a) Derive an expression for the torque experienced by an electric dipole kept in a uniformly electric field. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change ? Is the electric field due to a charge configuration with total charge zero, necessarily zero? The remaining faces of the cube given zero, (b) Graph between E Vs r Hence the net translating force on a dipole in a uniform electric field is zero. Question 6. Point is outside the spherical shell of radius ‘a’ but inside the spherical shell of radius ‘b’. The dipole moment of an electric dipole is a vector whose magnitude is either charge times the separation distance between the two opposite charges and the direction is along the dipole axis from the negative to the positive charge. Write its S.I. Answer: Using Gauss’s law, derive the expression for the electric field at a point Let σ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find \(\overrightarrow{\mathrm{E}}\) (b) Ratio of flux (b) For unstable equilibrium, the angle between p and E is 180°. Related questions . Two charges of magnitudes -3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. All India 2012) By symmetry, electric field E points outwards normal to the sheet. (i) the sheet is positively charged, Answer: (Delhi 2014) The direction of electric field is normal and inward to the surface. \(\phi_{\mathrm{E}}=\frac{q}{3 \varepsilon_{0}}\). This is the expression for electrostatic energy density in medium of dielectric constan K. In air of free space  (K = 1), therefore energy density, (a) Define electric flux. Question 19. Find the dimensions of (a) electric dipole moment p and (b) magnetic dipole moment M. The defining equations are p = q.d and M = IA; where d is distance, A is area, q is charge and I is current. Using this law derive an expression for the electric field due to a uniformly changed infinite plane sheet. Question 27. Electric field at a point on the surface of charged conductor, E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R^{2}}\) An electric dipole of dipole moment \(\overrightarrow { p } \) is placed in a uniform electric field \(\overrightarrow { E } \)?. (a) Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the centre of the dipole on the axial line. (All India 2010) By symmetry, electric field E points outwards normal to the sheet. (ii) Depict the orientation of the dipole in Two point charges 4 (J.C and +1 pC are separated by a distance of 2 m in air. Question 74. (a) Deduce the expression for the torque acting on a dipole of dipole moment \(\overrightarrow { p } \) in the presence of a uniform electric field E. S.I. Download the PDF Sample Papers Free for off line practice and view the Solutions online. (Delhi 2014) Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field. Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance 2a having dipole moment So net force on the dipole is zero Sample Papers. Answer: This indicates that electric field . If charge q gets larger, and the distance 2a gets smaller and smaller, keeping the product \(\overset{\rightarrow}{|P|}\) = q. (Outside Delhi 2017) So one can regard a line of force starting from a positive charge and ending on a negative charge. (Delhi 2012) Given a uniform electric field , find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane.

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