Poisson’s and Laplace’s Equation

Subject – Field Theory (2140909)
Branch – Electrical
Topic – Poisson’s and Laplace’s Equation

1. Gandhinagar Institute Of Technology Subject – Field Theory ( 2140909 ) Branch – Electrical Topic – Poisson’s and Laplace’s Equation

2. Name Enrollment No. Abhishek Chokshi 140120109005 Himal Desai 140120109008 Dheeraj Yadav 140120109010 Guided By – Prof. Supraja Ma’am

3. Laplace’s and Poisson’s Equation We have determined the electric field in a region using Coulomb’s law or Gauss law when the charge distribution is specified in the region or using the relation when the potential V is specified throughout the region. However, in practical cases, neither the charge distribution nor the potential distribution is specified only at some boundaries. These type of problems are known as electrostatic boundary value problems. For these type of problems, the field and the potential V are determined by using Poisson’s equation or Laplace’s equation. Laplace’s equation is the special case of Poisson’s equation.  

4. For the Linear material Poisson’s and Laplace’s equation can be easily derived from Gauss’s equation But, Putting the value of in Gauss Law, From homogeneous medium for which is a constant, we write Also, Then the previous equation becomes, Or,  

5. This equation is known as Poisson’s equation which state that the potential distribution in a region depend on the local charge distribution. In many boundary value problems, the charge distribution is involved on the surface of the conductor for which the free volume charge density is zero, i.e., ƍ=0. In that case, Poisson’s equation reduces to, This equation is known as Laplace’s equation.  

6. Laplace Equation in Three Coordinate System In Cartesian coordinates: and, Knowing Hence, Laplace’s equation is,  

7. In cylindrical coordinates : and, Knowing Hence, Laplace’s equation is,  

8. In spherical coordinates : and, Knowing Hence, Laplace’s equation is,  

9. Application of Laplace’s and Poisson’s Equation Using Laplace or Poisson’s equation we can obtain: Potential at any point in between two surface when potential at two surface are given. We can also obtain capacitance between these two surface.

10. EXAMPLE: Let and . Given point P(1,3,-1).Find V at point P. Also Find V satisfies Laplace equation. SOLUTION: V(1,3,-1) = 2*1* = -18 volt Laplace equation in Cartesian system is Differentiating given V, = = 0  

11. = = = = Adding double differentiating terms, = 0 + 4* + 12*x* *z Thus given V does not satisfy Laplace equation  

12. EXAMPLE: Consider two concentric spheres of radii a and b, a<b. The outer sphere is kept at a potential and the inner sphere at zero potential. Solve Laplace equation in spherical coordinates to find, The potential and electric field in the region between two spheres. Find the capacitance between them . SOLUTION: From the given data it is clear that V is a function of r only. i.e. and are zero. We know the Laplace's equation in spherical coordinate system as,  

13. This equation reduces to, By Integrating we get, Or, Integrating again we get, …………………( i )  

14. The given boundary condition are, Applying these boundary con to equ ( i ) …………(ii) Subtracting equ (i) from equ (ii) Putting value of A in equ (ii) we get,  

15. Putting these values in equ ( i ), we get potential V as, We have,  

16. (b) To Find capacitance : The flux density is, The charge density on the inner sphere i.e. at r=a is, The capacitance is now calculated by, …...Ans  

17. EXAMPLES: Two parallel conducting disks are separated by a distance 5mm at Z=0 and Z=5 mm. If V=0 at Z=0 and V=100 at z=5 mm. Find charge densities on the disks. SOLUTION: From the given con. It is clear that V is a fun. Of z only. Laplace’s equation of cylindrical coordinate system, we get Integrating twice we get,  

18. Where , A and B are constants of integration. To determine these constants apply the boundary conditions. V=0 at z=0 and V=100 Volts at z=5 mm Putting these in equ ( i ) Putting these in equ ( i ), the potential is obtained as, Now, the electric field between the circular disks is obtained using the relation,  

19. consider, the medium between the circular disks as free space, which gives the flux density as, Since D is constant between the phase and normal to the plates, we say that  

20. Uniqueness Theorem STATEMENT: A solution of P oisson’s equation (of which Laplace’s equation is a special case) that satisfies the given boundary condition is a unique solution. PROOF: Let us assume that we have two solution of Laplace’s equation, and , both general function of the coordinate use. Therefore, a nd  

21. Also assume that both and satisfies the same boundary condition . Let us define a new difference potential : Now i.e We have the vector identity, Let V= and D= , then we have From the above equation, , then we have  

22. Taking the volume integral over the volume gives, Now the divergent theorem is: The right hand side of above equation can now be written in term of surface integral as, The surface integral can be evaluated by considering the surface of very large sphere with radius R. when R is very large both and can be treated as a point charge, since size of it will be very small compared to R. Hence the surface integral on LHS decreases and approaches zero. Thus we have,  

23. Since is always positive, everywhere above equation is satisfied only if =0. If the gradient of i.e. is everywhere zero, then cannot change with any coordinates i.e. it has same value at all point on its boundary surface. Therefore, gives two identical solutions, i.e. unique solution  

24. Thank You