NOTE As, work done on a test charge by the electrostatic field due to any given charge configuration is independent of the path, hence potential difference is also same for any path. It happens due to the fact that no electric field exist inside a charged hollow conductor. Give reasons, Define an equipotential surface. (i) No, two equipotential surfaces cannot intersect each other because two normals can be drawn at intersecting point on two surfaces which give two directions of E at the same point which is impossible. 19.A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure. The potential due to a dipole at any point on equatorial line is zero, not an axial line. So, no work is done in moving a charge inside the shell. This is possible only when you have the best CBSE Class 12 Physics study material and a smart preparation plan. Candidates can click on the subject wise link to get the same. Therefore potential inside a charged conducting spherical shell equal to the potential at its surface. If q1, q2, q3 are charges on the plate of capacitors connected in parallel then total charge given by the source. Also calculate the electrostatic potential energy of the system of charges taking the value of charge, O = 2 X 10~7 C. [All India 2008] (ii) Closely spaced equipotential surfaces represent strong electric field and vice-versa. Ans. Ans. Electric field lines always begin on a positive charge and end on a negative charge and do not start or stop in mid space.

Its SI unit is J / C or volt and its dimension is [ML2T-3A-1]. Mock test are the practice test or you can say the blue print of the main exam. Another wire CD carrying 5 A is held directly above AB at a height of 1mm. 3.Electrostatic potential due to a point charge q at any point P lying at a distance r from it is given by With the help of Notes, candidates can plan their Strategy for particular weaker section of the subject and study hard. The electric potential at the surface of a charged conducting hollow sphere of radius 2 m is 500 volt. If V1, V2, V3,…. 31.Define the dipole moment of an electric dipole. V (rx) = potential at rx due to the external field It is a vector quantity and its direction is from negative charge towards positive charge. Is it a scalar or a vector quantity? The polar dielectric (like H2O, CO2, NH3 etc) are made up of polar atoms/molecules, in which the centre of positive charge does not coincide with the centre of negative charge of the atom. Ans. The relative closeness of the lines at some place give an idea about the intensity of electric field at that point.”. It is a vector quantity and its direction is in the direction of electrostatic force acting on positive charge. (ii) Draw schematically the equipotential surface corresponding to a field that uniformly increases in magnitude but remains constant in direction.

When two isolated charged conductors are connected to each other then charge is redistributed in the ratio of their capacitances. (ii)Potential Energy of a system of two charges in an external field Ans.Wrong, as potential due to an electric dipole is zero on equatorial line in spite of axial line. “An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point. The potential at the distance of 1.5 m from the centre is : (a) 375 V (b) 250 V (c) zero (d) 500V Answer: (d) 500V Electric potential inside and on surface of charged conducting sphere will be same. Ans. Work done is rotating an electric dipole in a uniform electric field from angle θ1 to θ2 is given by. Its SI unit is ‘coulomb-metre’ and its dimension is [LTA). When an electric dipole is placed in a non-uniform electric field, then a resultant force as well as a torque act on it. Ans. This causes a damage to the building. So, we can say that the potential difference between any two points on an equipotential surface is zero. Let us calculate the electric field at a point P that lies on the axis of the ring at distance x from its centre. How much work will be done in moving a test charge from point P{7,0,0) to Q (- 3, 0, 0)? where, r is radius of soap bubble and T is surface tension of soap bubble. 30.Two point charges 40, O are separated by lm in air.

In case of constant electric field along Z-direction,the perpendicular distance between equipotential surfaces remains same whereas for field of increasing magnitude, equipotential surfaces get closer as we go away from the origin.In both cases, surfaces be planes parallel to XY- plane. Electric lines of force are always perpendicular to an equipotential surface. (i)Calculate the potential difference between A and C. Electric potential energy of a charged conductor or a capacitor is given by. Ans. (i) Equipotential surface may be planer, solid etc. 6.Why there is no work done in moving a charge from one point to another on an equipotential surface? [Delhi 2008 C] 17.Two point charges q1 and q2 are located at q and r2, respectively in an external electric field E. Obtain the expression for the total work done in assembling this configuration. (iii) Electric potential at any point inside the conductor is constant and equal to potential. Is the work done by the field is moving a small positive charge from Q to P positive or negative?

Before appearing in the main examination, candidates must try mock test as it helps the students learn from their mistakes. Like charges repel and unlike charges attract each other. (vi) Electric Field Near an Infinite Plane Sheet of Charge, If infinite plane sheet has uniform thickness, then. Capacitance of an Isolated Spherical Conductor. At what point on the line joining the charges, is the electric field intensity zero? 10.Draw equipotential surfaces due to a single point charge. 15.What is the electric potential due to an electric dipole at an equatorial point?

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance exercises are given below to use it online or download in PDF form for offline. [All India 2009] Therefore, total potential energy of the system of two charges q•, and q2 placed at points with position vectors r, and r2 in the region of E is given by U= work done in bringing charge q from infinite to that position in E is equal to work done for charge q2 from infinite to that position in E +work done to that of charge q2 at these positions in presence of q1, 37. [All India 2011 c] Ans. (ii) Electric Field due to a Charged Spherical Shell, (iii) Electric Potential due to a Charged Conducting Spherical Shell. Ans. are potential differences across the plates of the capacitors then total voltage applied by thesource. V(r2) = potential at r2 due to the external field [All India 2008 C] (i)Potential Energy of a single charge in external field Potential energy of a single charge q at a point with position vector r, in an external field is qV(r),where V(r) is the potential at the point due to external electric field E.

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