Force on a charge due to charged sheet

[Yashbhatt]

Homework Statement

I want to find the force on a positive charge placed at a distance $d$ from a positively charged infinite plate. Of course, it can be simply done by finding the electric field due to the plate using Gauss's Law.

But my teacher suggested a different method and I am unable to comprehend it. He said that the field lines due to this particular configuration would be the same as the one in which a negative charge were to be placed on the opposite side of the plate at a distance $d$ forming a dipole.

Here $d = d_1$.

Homework Equations

$$F=\frac{kq_1q_2}{r^2}$$
$$E=\frac{\rho}{2\epsilon_0}$$
$$\ \ \ \oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$$
$$F=qE$$

The Attempt at a Solution

The normal electric field is $\frac{\rho}{2\epsilon_0}$ which gives a force independent from $d$ but using the above method, the force is $\frac{kQ^2}{(2d)^2}$ which depends on $d$ . What is the method at work? Is the force now simply the force between two positive and negative charges? i.e. Is the force on $+Q$ just $\frac{kQ^2}{(2d)^2}$ ?

 

[ehild]

Homework Statement

I want to find the force on a positive charge placed at a distance $d$ from a positively charged infinite plate. Of course, it can be simply done by finding the electric field due to the plate using Gauss's Law.

But my teacher suggested a different method and I am unable to comprehend it. He said that the field lines due to this particular configuration would be the same as the one in which a negative charge were to be placed on the opposite side of the plate at a distance $d$ forming a dipole.

Here $d = d_1$.

Homework Equations

$$F=\frac{kq_1q_2}{r^2}$$
$$E=\frac{\rho}{2\epsilon_0}$$
$$\ \ \ \oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$$
$$F=qE$$

The Attempt at a Solution

The normal electric field is $\frac{\rho}{2\epsilon_0}$ which gives a force independent from $d$ but using the above method, the force is $\frac{kQ^2}{(2d)^2}$ which depends on $d$ . What is the method at work? Is the force now simply the force between two positive and negative charges? i.e. Is the force on $+Q$ just $\frac{kQ^2}{(2d)^2}$ ?

Copy the original text of the problem, please.
You said that the infinite plate was positively charged. Was the plate metal or insulator? What was the surface charge density on it?

 

[Yashbhatt]

Copy the original text of the problem, please.
You said that the infinite plate was positively charged. Was the plate metal or insulator? What was the surface charge density on it?

The plate is a metal plate and the surface charge density is $\rho$.