Proving Charge Induced on Conductor is Equal to Sum of Imaginary Charges

In summary, the problem of finding the potential for a point charge in the vicinity of an infinite grounded sheet can be solved using the method of images. It can be proven using the Gauss law that the total charge induced on the sheet is equal to the sum of the imaginary charges, even for finite conductors. The potential in this scenario is azimuthally symmetric and can be expressed in cylindrical coordinates. The surface charge density and total charge on the grounded sheet can be calculated using the potential. Through integration, it can be shown that the total charge is always equal to the charge of the original point charge.
  • #1
harshant
30
0
Imagine that there is a point charge in vicinity of an infinite grounded sheet of conductor of arbitrary shape and size such that the problem of finding the potential can be solved by using the method of images. Is their a way to prove that the total charge induced on this sheet is always equal to the sum of the imaginary charges? (there exists a way for proving this for finite conductors using the Gauss law) I am confused because in all the texts I referred to this conclusion was stated to be 'expected' or 'obvious', and I couldn't see why it would be so.
 
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  • #2
Assume the grounded sheet is in the xy-plane and charge ##q## is at distance ##d## above it. The potential in region ##z>0## due to the charge and its image is azimuthally symmetric and expressed in cylindrical coordinates as
$$V(\rho,z)=\frac{q}{4\pi \epsilon_0}\left[\frac{1}{[\rho^2+(z-d)^2]^{1/2}}-\frac{1}{[\rho^2+(z+d)^2]^{1/2}}\right]$$
The surface charge density is given by
$$\sigma(\rho)=\epsilon_0~E_z(z=0)=- \epsilon_0 \left. \frac{\partial V(\rho,z)}{\partial z}\right|_{z=0}$$
$$\sigma(\rho)=-\frac{q}{4\pi \epsilon_0} \frac{2d}{(\rho^2+d^2)^{3/2}}$$
A ring of radius ##\rho## and width ##d\rho## bears charge ##dq=\sigma(\rho)2\pi \rho~ d\rho##. The total charge on the grounded sheet is
$$Q_{total}=\int_0^{\infty} \sigma(\rho)2\pi \rho~ d\rho$$
Do the integral and convince yourself that it is equal to ##-q##.
 

1. What is charge induction on a conductor?

Charge induction on a conductor is the process by which a charged object placed near a conductor causes a redistribution of charges on the surface of the conductor. This is due to the principle of like charges repelling each other and unlike charges attracting each other.

2. How does charge induction occur on a conductor?

Charge induction occurs when a charged object is brought near a conductor. The electrons in the conductor will be repelled by the like charge of the object and move to the opposite side of the conductor, leaving a positive charge on the side nearest to the object and a negative charge on the side farthest from the object.

3. What factors affect the amount of charge induced on a conductor?

The amount of charge induced on a conductor depends on the strength of the charged object, the distance between the object and the conductor, and the properties of the conductor, such as its shape and material. Conductors with larger surface areas will allow for more charge to be induced.

4. Can charge induction on a conductor be controlled?

Yes, charge induction on a conductor can be controlled by using a grounding wire. This wire connects the conductor to the ground, allowing any excess charge to flow away and neutralize the conductor. This is commonly used in electrical circuits to prevent static buildup and protect against electrical shocks.

5. How is charge induction on a conductor related to the concept of electric fields?

Charge induction on a conductor is directly related to electric fields, as the presence of a charged object creates an electric field that induces a redistribution of charges on the conductor. The strength and direction of the electric field will determine the amount and direction of charge induced on the conductor.

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