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
Wow, that's a shocking calculation you've got there! Let's zap through each part one by one:
(a) To find the electric field at the position of the point charge, we can use Gauss's Law and some fancy math, but I prefer to just say "it's in the air" and imagine tiny electric fairies dancing around it. Trust me, it's much more fun that way.
(b) The force experienced by the point charge due to the electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field. So, just plug in the values and voila! You'll get an answer that's undoubtedly electrifying.
(c) To find the potential difference at the position of the point charge, we can use the equation V = Ed, where V is the potential difference, E is the electric field, and d is the distance. Just remember that potential difference is like taking an electric rollercoaster ride, but without the loops and screams.
(d) The potential energy associated with the point charge can be found using the equation U = qV, where U is the potential energy, q is the charge, and V is the potential difference. It's like calculating how much energy is stored in an electric trampoline. Bounce, bounce!
(e) When the point charge is released, it experiences a force due to the electric field and undergoes acceleration using Newton's second law, F = ma. So, just rearrange the equation to find the acceleration, and you'll see the point charge zooming off like a lightning bolt!
Just remember, electricity can be shocking, so always look both ways before crossing an electric field!
To find the answers to these questions, we can use the principles of electrostatics. Let's go step by step:
(a) To find the electric field at the position of the point charge, we can use the formula for the electric field due to a uniformly charged plate at a point above it:
E = σ / (2ε₀),
where E is the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.
First, we need to determine the surface charge density (σ) by dividing the total charge (Q) by the area (A) of the plate:
σ = Q / A.
Given that the surface charge is +3.4 microcoulombs and the area is 4.0 m^2:
σ = 3.4 x 10^(-6) C / 4.0 m^2.
Now we can substitute the value of σ into the electric field formula to find E.
(b) The force experienced by the point charge can be found using the formula F = qE, where q is the charge of the point charge and E is the electric field strength.
(c) The potential difference at the position of the point charge can be calculated as the work done per unit charge, which is equal to the electric field multiplied by the distance. The formula is V = Ed.
(d) The potential energy associated with the point charge can be calculated using the formula PE = qV, where q is the charge and V is the potential difference.
(e) To calculate the acceleration of the point charge when released, we can use the formula F = ma, where F is the force experienced by the point charge and m is the mass of the point charge.
Now, let's calculate each of these values one by one.