poissons equation
TRANSCRIPT
AMIT KUMAR MOHAPATRA14MSL0005
POISSON’S EQUATION
Gauss Law In ElectrostaticsThe net electric flux through any closed surface is equal to 1⁄ε times the total electric charge enclosed within that closed surface.
0Q
E
ΦE is the electric flux through a closed surface S enclosing any volume V, Q is the total charge enclosed within S.
The electric flux ΦE is defined as a surface integral of the electric field:
S
E dSE.0
.QdSE
S
The Equation can be written in differential form as:
0
.
E
The Electric field in terms of potential can be defined as:
VE .
From above two equations
0
2
V
The above equation is called Poisson’s equation.
• Current-carrying components in high-voltage power equipment can be cooled to carry away the heat caused by ohmic losses. A means of pumping is based on the force transmitted to cooling fluid by charges in an electric field. The region between the electrodes contains a uniform charge ρo, which is generated at the left electrode and collected at the right electrode. Calculate the pressure of the pump if ρo =2.5 mC/m3 and Vo = 22 kV.
Consider a pump of length ‘l’ and area ‘a’. It has two electrode; left electrode is +V0 and right electrode is V where we have to collect the uniform charge.
+V0 0V
0 d x
• Given 0 = 2.5 mC/ and V0 = 22 kV
Since 0 0, we have to apply Poisson’s equation
As charge is generated at the left electrode and collected at right electrode that means charge travels along X-axis. So the potential V is depends on x. From boundary conditions V(x=0)=V0 and V(x=d)=0, then
3m
0
2
V
Integrating twice, we have
BAxxV
2
20
0
2
2
dxVd
where A and B are integration constants to be determined by applying the boundary condition.
when when The electric field is given by
Electric field is defined as the force per unit charge, so force is given by
0,0 VVx 0VB
dVdA 00
2
dx
dVAx
dxdVVE 000
0, Vdx
EdVF
0000 aVEdxdsFd
x
333
00
551022105.2mNP
VaFP
Pressure is defined as the force per unit area applied in a direction perpendicular to the surface.Therefore,
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