Derivation of Maxwell's stress and electrostatic attraction using Maxwell's equations
Problem Statement
Derivation of Maxwell's stress and electrostatic attraction using Maxwell's equations
Problem Statement
Two perfectly conductive plates of sufficiently large area are placed parallel to each other at a distance d in a vacuum. A DC power supply is connected between the two conductive plates, providing a constant potential difference V between the plates. The space between the plates is filled with vacuum (permittance ε0).
Let S be the area of the conductive plates, and assume that all edge effects are negligible. Also, assume that the electric field vector E=(Ex,Ey,Ez) between the plates is uniform and points in the direction normal to the plates (in the x axis direction).
From Maxwell's equations and general theory of electromagnetism, each component Tij (i,j∈{x,y,z}) of Maxwell's stress tensor T in a vacuum is defined using the electric field vector as follows:
Tij=ε0(EiEj−21δij∣E∣2) Here, δij represents the Kronecker delta.
The magnitude of the electrostatic attractive force F exerted by one conducting plate on another in the x axis can be calculated exactly by integrating the normal component of Maxwell's stress tensor over the closed surface A surrounding the plates, as follows: F=∮A∑jTxjnjdR Here, n=(nx,ny,nz) is the outward unit normal vector of the closed surface A.
Using this tensor calculation, express the normal component Txx in terms of ε0,V,andd, and derive an algebraic expression F representing the magnitude of the total electrostatic attractive force acting on one of the plates.
Constraints
- ε0=8.85×10−12 F/m
- S=0.500 m2
- V=3000 V
- d=2.00×10−3 m
Input Format
Calculate the magnitude of the electrostatic attractive force F [N] derived to A×10−6 N. Find the value of the coefficient A in this case, and input a natural number representing that value.
Solution
Explanation
Calculation of Maxwell's Stress Tensor Components
From the problem setting, the uniform electric field between the plates has a component only in the x axis direction. That is, the electric field vector is expressed as follows: E=(Ex,0,0) Since the potential difference is V and the plate spacing is d, its magnitude Ex is as follows: Ex=dV
We calculate the normal component Txx in the x axis direction of Maxwell's stress tensor from the definition. When i=x,j=x, δxx=1 and ∣E∣2=Ex2, so we obtain the following equation.
Txx=ε0(ExEx−21δxx∣E∣2)=ε0(Ex2−21Ex2)=21ε0Ex2 Txx=21ε0(dV)2
For the other components, when j=x, Ey=0,Ez=0 and δxj=0, therefore Txy=0,Txz=0.
Calculation of Force Using Closed Surface Integrals
Consider a thin rectangular closed surface A that encloses the entire electrode plate. On the surface facing the inside of the electrode plate (the space where the electric field exists), the outward unit normal vector is n=(1,0,0) or n=(−1,0,0). On the back side of the electrode plate (inside or outside the conductor), the electric field is 0, so the component of the stress tensor is also 0. Furthermore, the side integral does not contribute because it ignores edge effects.
Therefore, only the contribution from the surface where the electric field exists (area S) remains, and the force F in the x axis direction can be calculated by performing a surface integral in the form of a simple product, as follows:
F=Txx×S=21ε0S(dV)2
This result is in exact agreement with the equation for the inter-plate attractive force of a parallel-plate capacitor derived from the energy method (derivative of electrostatic energy with respect to virtual displacement).
Numerical Substitution and Calculation
Substitute the given numerical constraints. Ex=dV=2.00×10−33000=1.50×106 V/m Ex2=(1.50×106)2=2.25×1012 V2/m2
Use these to calculate the force F. F=21×(8.85×10−12)×0.500×(2.25×1012) F=21×0.500×8.85×2.25×10−12×1012 F=0.250×19.9125=4.978125 N
The input format is A×10−6 N, so we match the units and number of decimal places.
F=4.978125 N=4978125×10−6 N
Therefore, the coefficient A is 4978125.
Answer: 4978125