Electrostatic potential and capacitance

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1. Electrostatic potential and Capacitance Basic Concepts

2. Electrostatic Potential energy Consider an electric field produced by a source charge Q at origin. Now we need to calculate the amount of work required to move a test charge q from point A to B assuming that it does not affect the electric field produced by Q. So as we know that electrostatic force is conservative, the amount of work done in bringing the charge from A to B is change in potential energy of the charge at A and B or gain in potential energy. So .   Q q A B

3. potential Energy at a point Now we can assume that the potential energy is zero at infinity. So, if we assume point A to be at infinity in the previous example then, = . So, potential energy of a charge q at any poin (here, B) is the amount of work done by external force in bringing charge q from infinity to that point.  

4. Electrostatic potential Now the electrostatic potential at a point, just like electric field intensity, is the work done by external force in bringing a unit positive charge from infinity to that point. It is denoted by V. So = (from the previous example)  

5. Potential due to a system of charges Here if we consider a system of charges with their relative position vectors . Then potential due to individual charges would be ; ; Now, according to superposition principle, the net potential due to system of charges   P          

6. Potential due to a point charge Here consider a point charge Q at the origin and assume bringing a unit charge from infinity to any point at position vector . Now at some intermediate point , the net electrostatic force = Now work done by charge to a small distance is .   Q A        

7. Here potential can be obtained by integrating the work done from to . So integrating the amount of work done, we get potential at A. So electrostactic potential at a point due to a point charge Q,  

8. Potential due to an electric dipole Here let the midpoint of the dipole be at origin. Now, from the figure, by cosine rule, we can say that and . So taking r common we can say that Now as r>>a, we only consider first order terms of r/a So and   q - q 2a θ r     o a a A

9. So and Applying binomial theoram and neglecting higher degree terms (as r>>a), we get and ( i ) Now according to superposition principle, net potential at A = potential at A due to q + potential at A due to –q. So . Now from ( i ) we can say that Now . So So the potential on the dipole axis would be and on the equatorial plane it would be zero.  

10. Equipotential surfaces The equipotential surfaces are surfaces at which the potential at every point on the surface is constant. Also for any charged configuration, the electric field is always perpendicular to equipotential surfaces. Other wise it would have a component on the equipotential surface, and hence the potential at the surface would not be constant. Thus electric field should be perpendicular to equipotential surface.

11. Relation between field and potential Consider moving a unit charge from an equipotential surface a to equipotential surface B separated by a distance . Let the change in potential in moving from surface A to B be δ V and the electric field perpendicular to the surface be E. So the work done against the field in taking a unit charge from surface A to B would be . Thus As the charge is moving in the direction of electric field, would be negative. So = -| | Thus |E| = .   δ l A B V V+ δ V E

12. The two important conclusions that can be drawn from here are:- Electric field is in the direction in which the potential decreases the steepest. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

13. Potential energy in an external field Potential energy of a single charge q in an external field E with position vector r will be qV (r). Now for the potential energy of system of two charges at points and in an electric field E we can assume first taking a charge from infinity to point and then taking charge from infinity to point . So work done against would be V( ). Now when we take charge from infinity to from the superposition principle of electric fieds work done would be equal to te summation of work done against both the electric field and the charge So for charge work done would be equal to V( ) + So net work done = V( ) + V( ) +  

14. Potential energy of a dipole in an electric field Here let us assume that in the external field the angle of the dipole with the electric field changes from to . So, as we know from the last chapter, Thus, work done by external torque in an electric field would be equal to Now assuming potential energy at we can say that   θ - qE qE -q q E a

15. Electrostatics of Conductors The conductors contain mobile charge carriers known as electrons. There are some important results related to conductors. Inside a conductor, electrostatic field is zero. At surface of a charged conductor, electrostatic field must be normal to the surface at every point. The interior of a conductor can have no excess charge in static situation. Electrostatic potential is constant throughout the conductor and its value is same as on its surface. Electric field at the surface of the charged conductor ( being surface charge density and being a unit vector normal to the surface in the outward direction. Electrostatic shielding – If we consider a conductor with a cavity, then the electrostatic field inside the cavity is zero. This is known as electrostatice shielding.  

16. Dielectrics and Polarization Dielectrics are non conducting substances and thus have no charge carriers. In an external field, a net dipole movement inside a dielectric is developed. The dipole moment per unit volume Is known as polarisation and is denoted by P. For linear isotropic dielectrics (those having polar molecules) . Here is the constant characteristic of the dielectric and is known as susceptibility.  

17. Capacitors and Capacitance A capacitor is a system of two conductors separated by an insulator. Generally the conductors have charges Q and –Q and potential and respectively. The charge Q is called te charge of the capacitor. Now if we consider V as work done per unit charge in taking it from a plate with charge –Q to the plate with charge Q (potential difference). Then, V is found to be proportional to Q and the ratio of Q/V is a constant C = Q/V. Here C is known as the capacitance of the capacitor. Its unit is faraday and is denoted by F.  

18. The capacitance of a capcitor also depends upon the insulator (dielectric) separating the two conductors. Now if the electric potential between two conductors is very high, there is an increase in the electric field strength between the conductors. This ionizes the surrounding molecules of the insulator and accelerates the charges produced to oppositely charged plates, partly neutralizing the capacitor plates. In other words, we can say that the charge of the capacitor leaks away due to reduction in insulating power of the intervening dielectric. So the maximum electric field that a capacitor can withstand without break down (of its insulation property) is known as dielectric strength of the capacitor For air the dielectric strength is equivalent to .  

19. The parallel plate capacitor Here the electric field in the inner region is Now as we know , V = Ed, so   + + + + + + + - - - - - - - - Surface charge density σ Surface charge density - σ d Area A E

20. Effect of dielectric on Capacitance Here let the charge density due to polarisation in dielectric be So the net surface charge density on each plates would be Thus the net electric field would be and . For linear dielectrics is proportional to σ and we can write So And Here is permitivity of medium. So where is the capacitance of vaccum .   + + + + + + + - - - - - - - - Surface charge density σ Surface charge density - σ d Area A E - - - - - - - + + + + + +    

21. Combination of capacitors Capacitors in series Here the charge on plates of each capacitor would be This is in accordance with the fact that the charge would flow through the electric conductor connecting the two capacitors until charges on left plate of one capacitor and on right plate of other capacitor do not sum up to zero. Now let the voltage across the capacitors be . So net voltage . Dividing both the sides by Q we get (as C = Q/V)           + + + + + + - - - - - - + + + + + + - - - - - - + + + + + + - - - - - - + + + + + + - - - - - - Q Q Q Q

22. Capacitors in parallel Here the voltage of the capacitors remains same, say V. Assuming that the charge on the capacitors be net charge Dividing both the sides of the equation by V, we get     + + + + + + - - - - - -     + + + + + + - - - - - -     + + + + + + - - - - - -   V

23. Energy stored in a capacitor Here consider taking charge from one plate to another with the help of an external force. So the potenial across the capacitors at this time would be . Here net amount of work done Now integrating the above equation from charge 0 to charge we get So potential energy Now we can write As we know , Since Ad is the volume of the capacitor, u = U/Ad is the energy density of the capacitor. So energy density of a capacitor is   + + + + + + + - - - - - - -       +

24. Electric potential due to special charge distribution Point charge Here electric field at point P is V where   E r q   P

25. Charged conducting sphere for point outside the sphere ; r=R for point at the surface for point inside the sphere   E R   A B C R + + + + + + + + + +

26. Uniformly charged non conducting sphere for point outside the sphere ; r=R for point at the surface for point inside the sphere   E R   A B C R + + + + + + + + + + +  

27. Charged circular ring at an axial point   x R P + + + + + + + + + + + + r   V R

28. Due to charged circular disk   x R P + + + + + + + + + + + + r   E

29. Van de Graff Generator Principle:- Consider putting a charge Q on the surface of conducting spherical shell of radius R. So the potential due to charge Q anywhere inside the shell or outside it would be . Consider a conducting sphere of radius r and charge q placed inside the sphere of radius r. So the potential on the smaller sphere would be And on the surface of shell it would be Now the potential difference Now as q is positive, the potential difference between the smaller and larger sphere would always be positive. Thus if we connect the smaller and larger sphere by a conducting wire, charge would always flow from smaller to larger sphere however large may be the charge Q on the larger sphere. This is used to accumulate large amounts of charge on the outermost surface of the sphere in V an D e Graff generator.   q Q + + + + + + + + + + + + + + + + + + + + + + A B

30. src :- NCERT Capable of generating the potential differences of about 6 to 8 million volts A B

31. For any queries contact :- Lekha Periwal : [phone] Heer Mehta : [phone]