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## Main Question or Discussion Point

I have a question about calculating induced surface charge:

Consider a system composed of an infinite grounded conducting sheet located at z=0 and a point charge positioned some distance d above the sheet.

Via method of images, we can find the potential in the region z>0. When we are done, we will have V(x,y,z) for z>0.

Then, we typically proceed to find the induced surface charge on the conducting plane. We do so by noting that, immediately above the plane, the field only has a normal component, which is [tex] \sigma / \epsilon_0 \hat{n} [/tex].

Since

[tex]E= -\nabla V[/tex]

and E only has a component in the normal direction (i.e. the z direction)

[tex]E = -\frac{\partial V}{\partial z} \hat{n}[/tex]

and so

[tex]\sigma= -\epsilon_0 \frac{\partial V}{\partial z}[/tex]

where the partial derivative is evaluated at (or immediately above) z=0. We can easily evaluate this derivative since we know V(x,y,z) (from the method of images above).

However, this process of obtaining [tex]\sigma[/tex] assumes that the field directly below the conducting sheet is zero--this assumption is present in that we had to use the fact that the field directly above the plane is [tex] \sigma / \epsilon_0 \hat{n} [/tex] and we can only argue that that is true if E=0 below the sheet (the argument would go as follows: We know E always undergoes a discontinuity of [tex]\sigma/ \epsilon_0[/tex] across any surface boundary. Since E is zero below the infinite sheet, the entire discontinuity must be due to the field above the sheet, and hence the field above the sheet is [tex]\sigma / \epsilon_0[/tex]).

But how do we know that the field below the sheet is zero? If it is because the sheet shields the region below the plane from static fields, how do we know that it shield's the region?

Consider a system composed of an infinite grounded conducting sheet located at z=0 and a point charge positioned some distance d above the sheet.

Via method of images, we can find the potential in the region z>0. When we are done, we will have V(x,y,z) for z>0.

Then, we typically proceed to find the induced surface charge on the conducting plane. We do so by noting that, immediately above the plane, the field only has a normal component, which is [tex] \sigma / \epsilon_0 \hat{n} [/tex].

Since

[tex]E= -\nabla V[/tex]

and E only has a component in the normal direction (i.e. the z direction)

[tex]E = -\frac{\partial V}{\partial z} \hat{n}[/tex]

and so

[tex]\sigma= -\epsilon_0 \frac{\partial V}{\partial z}[/tex]

where the partial derivative is evaluated at (or immediately above) z=0. We can easily evaluate this derivative since we know V(x,y,z) (from the method of images above).

However, this process of obtaining [tex]\sigma[/tex] assumes that the field directly below the conducting sheet is zero--this assumption is present in that we had to use the fact that the field directly above the plane is [tex] \sigma / \epsilon_0 \hat{n} [/tex] and we can only argue that that is true if E=0 below the sheet (the argument would go as follows: We know E always undergoes a discontinuity of [tex]\sigma/ \epsilon_0[/tex] across any surface boundary. Since E is zero below the infinite sheet, the entire discontinuity must be due to the field above the sheet, and hence the field above the sheet is [tex]\sigma / \epsilon_0[/tex]).

But how do we know that the field below the sheet is zero? If it is because the sheet shields the region below the plane from static fields, how do we know that it shield's the region?

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