Question

A point charge q is brought to a position a distant d away from an infinite plane conductor held at zero potential. Use the method of images, to find:
i. The surface charge density induced on the plane (6 marks)
ii. The force between the plane and the charge by using Coulomb’s law for the force between the charge and its image (2 marks)
iii. The total force acting on the plane by integrating

Answer 1

Answer on Question #71428, Physics / Electromagnetism

A point charge $q$ is brought to a position a distant $d$ away from an infinite plane conductor held at zero potential. Use the method of images, to find:

i. The surface charge density induced on the plane;

ii. The force between the plane and the charge by using Coulomb's law for the force between the charge and its image;

iii. The total force acting on the plane by integrating.

Solution:

i.

$$\Phi_{+}(\vec{x}) = \frac{1}{4\pi\varepsilon_0} \frac{q}{|\vec{x} - \vec{x}''|} = \frac{q}{4\pi\varepsilon_0 \sqrt{(x^2 + y^2 + (z - d)^2}}$$ $$\Phi_{-}(\vec{x}) = \frac{1}{4\pi\varepsilon_0} \frac{-q}{|\vec{x} - \vec{x}''|} = \frac{-q}{4\pi\varepsilon_0 \sqrt{(x^2 + y^2 + (z + d)^2}}$$ $$\Phi(\vec{x}) = \frac{q}{4\pi\varepsilon_0 \left[ \sqrt{(x^2 + y^2 + (z - d)^2} - \sqrt{(x^2 + y^2 + (z + d)^2} \right]}$$ $$\sigma = -\varepsilon_0 \frac{\partial \Phi}{\partial z} \Bigg|_{z=0} = \varepsilon_0 \frac{q}{4\pi\varepsilon_0} \left[ \frac{(z - d)}{((x^2 + y^2 + (z - d)^2)^{3/2}} - \frac{(z + d)}{((x^2 + y^2 + (z + d)^2)^{3/2}} \right]$$ $$\sigma = -\frac{qd}{2\pi} \left( (x^2 + y^2 + d^2)^{-3/2} \right) = -\frac{qd}{2\pi} \left( (r^2 + d^2)^{-3/2} \right)$$

ii.

$$\vec{F} = q \vec{F} = q \frac{-q}{4\pi\varepsilon_0 (2d)^2} = -\frac{q^2}{16\pi\varepsilon_0 d^2} \vec{z}$$

iii.

$$F = \int \frac{\sigma^2}{2\varepsilon_0} dA = \int \frac{\left(\frac{qd}{2\pi}\right)^2 (r^2 + d^2)^{-3}}{2\varepsilon_0} dA = \int_0^\infty \int_0^{2\pi} r dr d\varphi \frac{\left(\frac{qd}{2\pi}\right)^2 (r^2 + d^2)^{-3}}{2\varepsilon_0}$$ $$F = 2\pi \frac{\left(\frac{qd}{2\pi}\right)^2}{2\varepsilon_0} \int^\infty r dr (r^2 + d^2)^{-3}$$ $$u = r^2 + d^2$$ $$dr = \frac{du}{2r}$$ $$F = 2\pi \frac{\left(\frac{qd}{2\pi}\right)^2}{2\varepsilon_0} \int r u^{-3} \frac{du}{2r} = 2\pi \frac{\left(\frac{qd}{2\pi}\right)^2}{2\varepsilon_0} \frac{1}{2} \int du u^{-3} = 2\pi \frac{\left(\frac{qd}{2\pi}\right)^2}{2\varepsilon_0} \frac{1}{2} \left[ -\frac{1}{2} (r^2 + d^2)^{-2} \right] \Bigg|_0^\infty = \frac{q^2}{16\pi\varepsilon_0 d^2}$$

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