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In principle, you have to find the solution of the Laplace equation in the half-space of the point charge whichwhich fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0$V=0$ plane, where however, it doesn't fulfill the boundary V=0$V=0$ condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem, which has been proven mathematically to be unique.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0$V=0$ is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem, which has been proven mathematically to be unique.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the $V=0$ plane, where however, it doesn't fulfill the boundary $V=0$ condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem, which has been proven mathematically to be unique.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition $V=0$ is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

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In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem, which has been proven mathematically to be unique.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem, which has been proven mathematically to be unique.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

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In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

In principle, you have to find the solution of the Laplace equation in the half-space of the point charge which fulfills the boundary condition for the potential zero on the grounded plane. The image method makes this easy in this case. The Coulomb potential of the point charge is a solution of the Laplace equation everywhere in infinite space (except at the point charge location). In particular, also at the V=0 plane, where however, it doesn't fulfill the boundary V=0 condition. The Laplace equation is linear, and if you find a combination of solutions that fulfills the boundary condition, you have found a solution of the boundary value problem.

The trick here is that you introduce a virtual, oppositely charged point charge at a mirrored distance on the other side of the surface, which also gives a Coulomb potential solution of the Laplace equation, but of opposite sign. Then the potentials of the original and the mirror charge add to zero at the surface, so that the boundary condition V=0 is fulfilled there. Thus you have found a solution of the boundary value problem by just adding the real and the virtual point charge Coulomb potentials. You have to be aware though, that this potential solution is only valid in the half-space of the original point charge including the plane surface.

Note: It should be clear that, once you have the potential solution, you obtain the electric field as a gradient of the potential.

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