- #1
Yashbhatt
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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}## ?