A large conducting plate of area equal to 4.0 m^2 has an evenly distributed surface charge of + 3.4 micro coulomb on its surface.A point charge of mass equal to 5.3 x 10^-6 kg and carrying a charge of 1.2 nC is held at 10 cm above the center of the plate.Determine the following :

(a) electric field at position of the point charge (b) the force experienced by the point charge as a result of the electric field (c) potential difference at the position of the point charge (d) potential energy associated with the point charge (e) acceleration of the point charge when released

To find the required values, we can use the principles of electrostatics and Coulomb's law. Here's how we can obtain the answers:

(a) To determine the electric field at the position of the point charge, we can use Gauss's law. Since the large conducting plate is evenly distributed with a surface charge, the electric field due to the plate will be zero inside it. Therefore, we only need to consider the point charge.

The electric field at a distance r from a point charge is given by:
E = k * (q / r^2)
where E is the electric field, k is the electrostatic constant (= 8.99 x 10^9 Nm^2/C^2), q is the charge of the point charge, and r is the distance from the charge.

In this case, q = 1.2 nC = 1.2 x 10^-9 C and r = 10 cm = 0.1 m.
Substituting these values into the equation, we get:
E = (8.99 x 10^9 Nm^2/C^2) * (1.2 x 10^-9 C / (0.1 m)^2)

Calculating this expression will give us the electric field at the position of the point charge.

(b) The force experienced by the point charge as a result of the electric field can be calculated using the formula:
F = q * E
where F is the force, q is the charge of the point charge, and E is the electric field.

Substituting the given values into the equation will give us the force experienced by the point charge.

(c) The potential difference at the position of the point charge can be calculated using the formula:
V = k * (Q / r)
where V is the potential difference, k is the electrostatic constant, Q is the surface charge of the conducting plate, and r is the distance between the point charge and the plate.

In this case, Q = 3.4 μC = 3.4 x 10^-6 C and r = 0.1 m.
Substituting these values into the equation will give us the potential difference at the position of the point charge.

(d) The potential energy associated with the point charge can be calculated using the formula:
U = q * V
where U is the potential energy, q is the charge of the point charge, and V is the potential difference.

Substituting the given values into the equation will give us the potential energy associated with the point charge.

(e) The acceleration of the point charge when released can be calculated using Newton's second law, which states that:
F = m * a
where F is the force acting on the point charge, m is the mass of the point charge, and a is the acceleration.

In this case, we already calculated the force experienced by the point charge (from step (b)) and the mass of the point charge is given as 5.3 x 10^-6 kg. Substituting these values into the equation will give us the acceleration of the point charge when released.

By following these steps and performing the necessary calculations, you should be able to find the answers to all the given questions.