Method of images

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Method of images
관련코스 전기역학
소분류 물리
선행 키워드 Laplace's equation
연관 키워드

Charges near conducting plane

Point charge

Let us use cylindrical coordinates so that a point charge sits at (0,0,a). Then, an image charge will sit at (0,0,-a) and the potential function will be given as

[math]V\left(\rho,\varphi,z\right) = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{\sqrt{\rho^2 + \left(z-a \right)^2}} + \frac{-q}{\sqrt{\rho^2 + \left(z+a \right)^2}} \right) \,[/math]

Then, the surface charge density is:

[math]\sigma = -\epsilon_0 \frac{\partial V}{\partial z} \Bigg|_{z=0} = \frac{-q a}{2 \pi \left(\rho^2 + a^2\right)^{3/2} }[/math]

whose total integration equals -q as expected.

[math] \begin{align} Q_t & = \int_0^{2\pi}\int_0^\infty \sigma\left(\rho\right)\, \rho\,d \rho\,d\theta \\[6pt] & = \frac{-qa}{2\pi} \int_0^{2\pi}d\theta \int_0^\infty \frac{\rho\,d \rho}{\left(\rho^2 + a^2\right)^{3/2}} \\[6pt] & = -q \end{align} [/math]

Force and energy

The force the charge experiences will be the same as the force from image charge:


and the potential energy will be given as a line integral of this force from infinity to current position:

[math]W = -\int_{\infty}^d \mathbf{F}\cdot d\mathbf{l} = \int_{\infty}^{d} \frac1{4\pi\epsilon_0}\frac{q^2}{4z^2}dz = -\frac1{4\pi\epsilon_0} \frac{q^2}{4d}[/math]

Electric dipole

The image of an electric dipole moment p at [math](0,0,a)[/math] above an infinite grounded conducting plane in the xy-plane is a dipole moment at [math](0,0,-a)[/math] with equal magnitude and direction rotated azimuthally by π. That is, a dipole moment with Cartesian components [math](p\sin\theta\cos\phi,p\sin\theta\sin\phi,p\cos\theta)[/math] will have in image dipole moment [math](-p\sin\theta\cos\phi,-p\sin\theta\sin\phi,p\cos\theta)[/math].

The dipole experiences a force in the z direction, given by


and a torque in the plane perpendicular to the dipole and the conducting plane,

[math]\tau=-\frac{1}{4\pi\epsilon_0}\frac{p^2}{16a^3}\sin 2\theta[/math]

Charges near conducting sphere