Electrostatics. Ans.The particles 1 and 2 are negatively charged and particle 3 is positively charged.

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Georgia Gibbs
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1 Electrostatics [ Two marks each] Q1.An electric dipole with dipole moment C m is aligned at 30 with the direction of a uniform electric field of magnitude N C 1. Calculate the net force and magnitude of the torque acting on the dipole. Ans1. Electric dipole moment, p = C m Angle made by p with a uniform electric field, θ = 30 Electric field, E = N C 1 net force on the dipole = qe + (-qe) =0 Torque acting on the dipole is given by the relation, pe sinθ Therefore, the magnitude of the torque acting on the dipole is 10 4 N m. Q2. Fig shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio? Ans.The particles 1 and 2 are negatively charged and particle 3 is positively charged. The charge to mass ratio (e/m) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio. Q3. Define equipotential surface. Draw the diagram to show the equipotential surface of an electric dipole. Ans. : It is the surface at every point of which the potential is same.

[ Three marks each] Q4. State Gauss theorem in electrostatics. A thin charged wire of infinite length has line charge density λ. Derive expression for electric field at a distance r.

2 [ Three marks each] Q4. State Gauss theorem in electrostatics. A thin charged wire of infinite length has line charge density λ. Derive expression for electric field at a distance r. Ans: Gauss s Law: Electric flux through a closed surface is 1/ε 0 times the net charge enclosed by it. i.e. To calculate the field at P we consider a cylindrical Gaussian surface with wire as axis, radius r and length l as shown in the figure. The electric lines of force are parallel to the end faces of the cylinder and hence the component of the field along the normal to the end faces is zero. The field is radial everywhere and hence the electric flux crosses only through the curved surface of the cylinder. If E is the electric field intensity at P, then the electric flux through the Gaussian surface is According to gauss theorem electric flux is hence Q5. Derive the expression for energy stored in a capacitor and hence obtain the expression for energy density of a parallel plate capacitor Ans: Consider a parallel plate capacitor of capacitance C. Let at any instant the charge on the capacitor be Q. Then potential difference between the plates will be V = Q /C Suppose the charge on the plates increases by dq. The work done will be The total work done is * + This work done is stored as electrical potential energy. * +

3 Energy density of parallel plate capacitor: Capacitance of a parallel plate capacitor is hence energy density * + Q6: A parallel plate capacitor is charged by a battery to a potential V. It is disconnected and a dielectric slab is inserted to completely fill the space between the plates. How will i) its capacitance ii) electric field between the plates and iii) energy stored in the capacitor be affected? Justify your answer in each case. Ans: i) On inserting dielectric capacity of the capacitor increases ii)electric field decreases iii) As the battery is disconnected charge on the capacitor remains constant. Hence the energy stored in the capacitor is given by As C increases therefore U decreases [ Five marks each] Q7. Charge q is distributed uniformly on a spherical shell of radius R. Using Gauss law derive expression of electric field at a distance r from the centre when (i)r>r (ii) r=r (iii) r<r Draw a graph to the variation of E with r. Ans: Consider a hollow conducting sphere of radius R with its centre at O. let σ be its surface density Let r be the distance between the centre of the spherical shell and the point where E is to be calculated. Case (i)r>r At points outside the sphere the electric field is radial every where because of spherical symmetry. Total electric flux According to Gauss theorem electric flux is hence

4 * + Electric field due to charged shell is same as that due to a point charge q placed at the centre of shell. Case (i)r=r When point P lies on the surface of the shell or sphere, r = R hence [ ] Case (i)r<r The gaussian surface does not enclose any charge, (charge resides on the surface of the shell) E = 0 E 0 r<r R r>r Q8. What is capacitance? State its S I unit and dimensions. Derive expression for capacity of parallel plate capacitor with a dielectric slab of thickness t (t<d) between the plates. Ans: Capacitance of a capacitor is equal to the ratio of magnitude of charge q (on either conductor) to the potential difference V between the conductors. i.e. C it is measured in farad. [M -1 L -2 T 4 A 2 ] Let the surface charge density on the plates be if E 0 is electric field in air and E i is electric field in dielectric,the Potential difference between the plates is given by +σ t -σ Capacity of a capacitor d

5 Q9. Define electric dipole moment. State it SI unit and dimensions. Derive expression for electric field at a point on the equatorial line of dipole. Ans: Electric dipole moment is a vector whose magnitude is equal to the product magnitude of either charge q and the separation between them & is directed from negative to positive charge. i.e. p = q x 2a p where p is the unit vector directed from q to +q its SI unit is Cm. [LTA] Consider a dipole consisting of -q and +q separated by a distance 2a. Let P be a point Consider a point P on the equatorial line. The resultant intensity is the vector sum of the intensities along PA and PB. E A and E B can be resolved into vertical and horizontal components. The vertical components of E A Sinθ and E B Sinθ cancel each other as they are equal and oppositely directed. It is the horizontal components which add up to give the resultant field. E = 2E A cos

6 using p= 2aq.for a<<r [ ] E= Q10. a) You are given an air filled parallel plate capacitor C 1. The space between its plates is now filled with the slabs of dielectric constant K 1 and K 2 as shown in C 2. Find the capacitance of the capacitor C 2 if area of the plates is A, distance between the plates is d. After introduction of dielectric capacitor acts as if two capacitors each of area and separation d/2 are connected in series b) State the formula for potential energy of an electric dipole placed in a uniform electric field at angle Ɵ. In which orientation a dipole placed in a uniform electric field is in i)stable, ii) Unstable Equilibrium? Ans: formula for potential energy [ ] (a) For stable equilibrium the angle between p and E must be 0 0 (b) For unstable equilibrium the angle between p and E must be *********************************

Chapter 4. Electrostatic Fields in Matter

Chapter 4. Electrostatic Fields in Matter 4.1. Polarization A neutral atom, placed in an external electric field, will experience no net force. However, even though the atom as a whole is neutral, the