Here's an alternative derivation of Eq. 3.10 (the surface charge density

induced on a grounded conducted plane by a point charge qa distance dabove

the plane). This approach (which generalizes to many other problems) does not

rely on the method of images. The total field is due in part to q,and in part to the

induced surface charge. Write down the zcomponents of these fields-in terms of

qand the as-yet-unknown σ(x,y)-just below the surface. The sum must be zero,

of course, because this is inside a conductor. Use that to determine σ.

Short Answer

Expert verified

The surface charge density induced on a grounded conducted plane by a point charge qa distance dabove the plane is -qd2πx2+y2+d232.

Step by step solution

01

Given data

There is a point charge qata distance dabove a grounded conducted plane.

02

 Step 2: Define electric field due to point charge and uniform surface charge density

The field due to a point charge q at a distance r from it is

Eq=14πε0qr3r^…… (1)

Here, ε0is the permittivity of free space.

The field due to a uniform plane surface charge density σ is:

Eσ=σ2ε0n^…… (2)

Here, n^is the unit vector perpendicular to the plane and pointing away from it.

03

Charge density induced on the plane

From equation (2), the z component of electric field on any point on the conducting plane due to the point charge q at a height d is

Eqz=-14πε0qdx2+y2+d23/2

Close to the surface, the induced surface charge can be considered constant. From equation (2) the field just below the surface is

Eσz=-σ2ε0

The field inside a conductor should be zero. Hence, the density is solved as:

-14πε0qdx2+y2+d23/2-σ2ε0=0σx,y=-qd2πx2+y2+d232

Thus, the induced charge density is -qd2πx2+y2+d232.

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Most popular questions from this chapter

In Section 3.1.4, I proved that the electrostatic potential at any point

in a charge-free region is equal to its average value over any spherical surface

(radius R )centered at .Here's an alternative argument that does not rely on Coulomb's law, only on Laplace's equation. We might as well set the origin at P .Let Vave(R)be the average; first show that

dVavedR=14πR2V.da

(note that the R2in da cancels the 1/R2out front, so the only dependence on R

is in itself). Now use the divergence theorem, and conclude that if Vsatisfies

Laplace's equation, then,Vave(0)=V(P),forallR18.

DeriveP3(x)from the Rodrigues formula, and check that P3(cosθ)satisfies the angular equation (3.60) for I=3. Check that P3and P1are orthogonal by explicit integration.

The potential at the surface of a sphere (radius R) is given by

V0=kcos3θ,

Where k is a constant. Find the potential inside and outside the sphere, as well as the surface charge density σ(θ)on the sphere. (Assume there's no charge inside or outside the sphere.)

A circular ring in thexy plane (radius R , centered at the origin) carries a uniform line charge λ. Find the first three terms(n=0,1,2) in the multi pole expansion for V(r,θ).

(a) Suppose the potential is a constant V0over the surface of the sphere. Use the results of Ex. 3.6 and Ex. 3.7 to find the potential inside and outside the sphere. (Of course, you know the answers in advance-this is just a consistency check on the method.)

(b) Find the potential inside and outside a spherical shell that carries a uniform surface charge σ0, using the results of Ex. 3.9.

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