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Electrostatics Tutorials
   ELECTRIC CHARGE BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTRIC CHARGE BY S.S.EDUCATION
   COULOMB’S LAW BY S.S.EDUCATION
   PROBLEM SOLVING TRICKS BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTROSTATICS FORCE BY S.S.EDUCATION
   ELECTROSTATIC FIELD BY S.S.EDUCATION
   ELECTROSTATIC LINES OF FORCE BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTRIC LINES OF FORCE BY S.S.EDUCATION
   ELECTRIC FIELD DUE TO A POINT CHARGE BY S.S.EDUCATION
   SUPERPOSITION OF ELECTRIC FIELDS BY S.S.EDUCATION
   ELECTRIC FIELD INTENSITY ON THE AXIS OF A UNIFORMLY CHARGED RING BY S.S.EDUCATION
   SPECIAL CASES BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTRIC FIELD BY S.S.EDUCATION
   ELECTRIC FLUX BY S.S.EDUCATION
   GAUSS LAW BY S.S.EDUCATION
   APPLICATION OF GAUSS’S LAW BY S.S.EDUCATION
   EXAMPLE BASED ON ELECTRIC FLUX BY S.S.EDUCATION
   ELECTRIC POTENTIAL AND POTENTIAL ENERGY BY S.S.EDUCATION
   ELECTRIC POTENTIAL BY S.S.EDUCATION
   POTENTIAL DIFFERENCE BY S.S.EDUCATION
   RELATION BETWEEN E AND ELECTRIC POTENTIAL BY S.S.EDUCATION
   ELECTRIC POTENTIAL DUE TO POINT CHARGES BY S.S.EDUCATION
   POTENTIAL DUE TO A CHARGED SPHERICAL SHELL BY S.S.EDUCATION
   POTENTIAL DUE TO A CHARGED CONDUCTING SPHERE BY S.S.EDUCATION
   POTENTIAL DUE TO A CHARGED NON-CONDUCTING SPHERE BY S.S.EDUCATION
   POTENTIAL DUE TO A RING AT A POINT LYING ON ITS AXIS BY S.S.EDUCATION
   EQUIPOTENTIAL SURFACE BY S.S.EDUCATION
   ELECTRIC STRENGTH BY S.S.EDUCATION
   ELECTRIC POTENTIAL ENERGY BY S.S.EDUCATION
   ELECTRON VOLT BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTRIC POTENTI AL BY S.S.EDUCATION
   EXAMPLES BASED ON ELECTRIC POTENTIAL ENERGY BY S.S.EDUCATION
   ELECTRIC DIPOLE BY S.S.EDUCATION
   ELECTRIC FIELD AT AN AXIAL POINT BY S.S.EDUCATION
   ELECTRIC FIELD AT AN EQUATORIAL POINT BY S.S.EDUCATION
   ELECTRIC FIELD BY S.S.EDUCATION
   FORCE AND TORQUE ON A DIPOLE PLACED BY S.S.EDUCATION
   ELECTRIC POTENTIAL DUE TO A DIPOLE BY S.S.EDUCATION
   POTENTIAL ENERGY OF A DIPOLE BY S.S.EDUCATION
   EXAMPLES BASED ON DIPOLE SYSTEM BY S.S.EDUCATION
   CHARGED LIQUID DROP BY S.S.EDUCATION
   EXAMPLE BASED ON CHARGED LIQUID DROP BY S.S.EDUCATION
   FORCE ON A CHARGED SURFACEBY S.S.EDUCATION
   EQUILIBRIUM OF A CHARGED SOAP BUBBLE BY S.S.EDUCATION
   MOTION OF A CHARGED PARTICLE IN ELECTRIC FIELD BY S.S.EDUCATION
   EXAMPLES BASED ON MOTION OF A CHARGED PARTICLE BY S.S.EDUCATION
   ELECTROSTATIC BEHAVIOUR OF CONDUCTORS BY S.S.EDUCATION
   INSULATORS BY S.S.EDUCATION
   ATMOSPHERIC ELECTRICITY BY S.S.EDUCATION
   POINTS TO REMEMBER BY S.S.EDUCATION
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EXAMPLES BASED ON ELECTROSTATICS FORCE

Example – 1 Atomic number of copper is 29, its atomic weight is 63.5 gm/mole. If two pieces of copper having weight 10 gm are taken and on one of the pieces 1 electron per 1000 atom is transferred to the other piece and there after this these pieces are placed 10 cm apart, then what will be the coulomb force between them.
Solution : In 1 mole copper (63.5 gm) there are 6 × 1023 atoms (NA = Avogadro number = 6 × 1023 atoms)
Number of atoms = = 9.45 x 1022
For every 1000 atoms, 1 electron is transferred therefore total number of transferred electron is

Therefore charge on one piece is 9.45 × 1019 e and on the other will be (9.45 × 1019e) Force when they are kept 10 cm apart


Example – 2 On an insulated rod having length L, Q charge is evenly distributed and a point charge q is placed at a distance d from one of its ends as shown. What will be the total electric force on q.
Solution : Considering a small element dx of the rod at a distance x from q, force on q due to this small element

where dQ is charge in small element dx and dQ = (Q/L) dx


Example – 3 Three equal charges (Q, each) are placed on the vertices of an equilateral triangle of side a. What is the resultant force on any one charge due to the other two?
Solution : The charges are shown in fig. The resultant force

From symmetry the direction is as shown along y-axis.
Example – 4 Five point charges, each of value +q are palced on five vertices of a regular hexagon of side L m. What is the magnitude of the force on a point charge of value -q coloumb placed at the centre of the hexagon?
Solution : If there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six charges on -q at O will be zero (as the forces due to individual charges will balance each other), i.e., ®FR = 0

Now If ®f is the force due to sixth charge and F® due to remaining five charges, ®F + ®f = 0

Example - 5 Two charged spheres of radius ‘R’ are kept at a distance ‘d’ (d > 2R). One has a charge +q and the other -q. The force between them will be-

[1] [2] [3] [4] None of these
Solution : [2] Redistribution of charge will take place due to mutual attraction and hence effective distance will be less than d.


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