A point charge \( q \) is placed at a distance \( d \) above an infinite, grounded conducting plate placed on the \( xy \)-plane at \( z = 0 \).
The electrostatic potential in the \( z > 0 \) region is given by \( \phi = \phi_1 + \phi_2 \), where:
\( \phi_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{x^2 + y^2 + (z - d)^2}} \)
\( \phi_2 = - \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{x^2 + y^2 + (z + d)^2}} \)
Which of the following option(s) is/are correct?
A point charge \( q \) is placed at a distance \( d \) above an infinite, grounded conducting plate placed on the \( xy \)-plane at \( z = 0 \).
The electrostatic potential in the \( z > 0 \) region is given by \( \phi = \phi_1 + \phi_2 \), where:
\( \phi_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{x^2 + y^2 + (z - d)^2}} \)
\( \phi_2 = - \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\sqrt{x^2 + y^2 + (z + d)^2}} \)
Which of the following option(s) is/are correct?
Show Hint
- The magnitude of the force experienced by the point charge \( q \) is \( \frac{1}{16 \pi \epsilon_0} \frac{q}{d^2} \)
- The electrostatic energy of the system is \( \frac{1}{8 \pi \epsilon_0} \frac{q^2}{d} \)
- The induced surface charge density on the plate is proportional to \( \frac{1}{\sqrt{x^2 + y^2 + d^2}} \)
- The electrostatic potential \( \phi_1 \) satisfies Poisson’s equation for \( z>0 \)
The Correct Option is D
Solution and Explanation
Step 1: The total electrostatic potential is the sum of the potentials due to the charge and the image charge. The image charge method is used to solve for the potential above the conducting plane.
Step 2: The force on the point charge \( q \) due to the image charge is given by Coulomb’s law. However, the question is focused on the potential and its properties, so the force magnitude and energy calculations are secondary and do not match the correct answer.
Step 3: The potential \( \phi_1 \) represents the potential of a charge above a grounded conducting plane, which satisfies Poisson’s equation for \( z > 0 \) since it corresponds to a solution of Laplace’s equation with boundary conditions at the plate.
Step 4: The potential \( \phi_2 \) represents the image charge, ensuring the boundary condition at the grounded conducting plane is satisfied.
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