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EEH
EEH
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This is more or less a restatement of Philip’s answer. Hopefully, in a more accessible way for you.

First, the lines radiating outward from q aren't field lines per sayse. They are actually a graphical representation of E's vector field along the axis.

Second, the vector field drawn is for $E_q$ only. It's easy to see that the magnitude of vector field is dropping off with distance prior to passing through the plane. As you point out if the surface is a conductor, the magnitude of the E-field would be zero at the plane (and on the other side due to shielding the effect of conductors).

First, the lines radiating outward from q aren't field lines per say. They are actually a graphical representation of E's vector field along the axis.

Second, the vector field drawn is for $E_q$ only. It's easy to see that the magnitude of vector field is dropping off with distance prior to passing through the plane. As you point out if the surface is a conductor, the magnitude of the E-field would be zero at the plane (and on the other side due to shielding the effect of conductors).

This is more or less a restatement of Philip’s answer. Hopefully, in a more accessible way for you.

First, the lines radiating outward from q aren't field lines per se. They are actually a graphical representation of E's vector field along the axis.

Second, the vector field drawn is for $E_q$ only. It's easy to see that the magnitude of vector field is dropping off with distance prior to passing through the plane. As you point out if the surface is a conductor, the magnitude of the E-field would be zero at the plane (and on the other side due to shielding the effect of conductors).

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EEH
EEH
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You don’t give us a ton to work off of. HoweverFirst, I am pretty confident the graphic is showing super position of field lines*. That is the field lines of E due toradiating outward from q are shown. The decreasing magnitude is due to the Earen't field decreasing over distance like $1/r^2$lines per say. You can tell this is the intent because theThey are actually a graphical representation of E's vector field lines magnitude drops before reachingalong the planeaxis.

Your last sentence tells me you think itSecond, the vector field drawn is a conductorfor $E_q$ only. If you knowIt's easy to see that the sheetmagnitude of vector field is a conductor as opposeddropping off with distance prior to a dielectric,passing through the above must be trueplane. We know this because asAs you saypoint out if the charge on either sidesurface is a conductor, the magnitude of the E-field would be zero at the plane will shield(and on the other side fromdue to shielding the point charge.

*note, these aren’t actually field lines. They are sketcheffect of vector field along axisconductors).

You don’t give us a ton to work off of. However, I am pretty confident the graphic is showing super position of field lines*. That is the field lines of E due to q are shown. The decreasing magnitude is due to the E field decreasing over distance like $1/r^2$. You can tell this is the intent because the field lines magnitude drops before reaching the plane.

Your last sentence tells me you think it is a conductor. If you know that the sheet is a conductor as opposed to a dielectric, the above must be true. We know this because as you say the charge on either side of the plane will shield the other side from the point charge.

*note, these aren’t actually field lines. They are sketch of vector field along axis.

First, the lines radiating outward from q aren't field lines per say. They are actually a graphical representation of E's vector field along the axis.

Second, the vector field drawn is for $E_q$ only. It's easy to see that the magnitude of vector field is dropping off with distance prior to passing through the plane. As you point out if the surface is a conductor, the magnitude of the E-field would be zero at the plane (and on the other side due to shielding the effect of conductors).

added 244 characters in body
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EEH
EEH
  • 91
  • 7

You don’t give us a ton to work off of. However, I am pretty sureconfident the graphic is showing super position of field lineslines*. That is the field lines of E due to q are shown. The decreasing magnitude is due to the E field decreasing over distance like $1/r^2$. You can tell this is the intent because the field lines magnitude drops before reaching the plane.

Your last sentence tells me you think it is a conductor. If you know that the sheet is a conductor as opposed to a dielectric, the above must be true. We know this because as you say the charge on either side of the plane will shield the other side from the point charge.

*note, these aren’t actually field lines. They are sketch of vector field along axis.

I am pretty sure the graphic is showing super position of field lines. That is the field lines of E due to q are shown. The decreasing magnitude is due to the E field decreasing over distance like $1/r^2$.

Your last sentence tells me you think it is a conductor. If you know that the sheet is a conductor as opposed to a dielectric, the above must be true. We know this because as you say the charge on either side of the plane will shield the other side from the point charge.

You don’t give us a ton to work off of. However, I am pretty confident the graphic is showing super position of field lines*. That is the field lines of E due to q are shown. The decreasing magnitude is due to the E field decreasing over distance like $1/r^2$. You can tell this is the intent because the field lines magnitude drops before reaching the plane.

Your last sentence tells me you think it is a conductor. If you know that the sheet is a conductor as opposed to a dielectric, the above must be true. We know this because as you say the charge on either side of the plane will shield the other side from the point charge.

*note, these aren’t actually field lines. They are sketch of vector field along axis.

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EEH
EEH
  • 91
  • 7