It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel . The potential energy is stored in the capacitor will be 2.7×10⁻⁵.
<h3>What is parallel plate capacitor ?</h3>It is an type capacitor is an in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates. helps to stop the flow of electric current through it due to its non-conductive nature .
The given data in the problem is;
Q is the charge= 1.5 µC
V is the change in voltage across the plates is = 36 V.
U is the potential energy=?
The formula for the potential energy is given by;
Hence the potential energy is stored in the capacitor will be 2.7×10⁻⁵.
To learn more about the parallel plate capacitor refer to the link;
The magnetic field along the central axis of the cylinder will be <em>determined </em>using the formula given that all parameters are known
Generally, A magnetic field is also referred to as a vector field that shows the magnetic presence on electric charges and magnetic materials.
The formula for magnetic field is
Where
Therefore, the magnetic field along the central axis of the cylinder will be determined using the formula given that all parameters are known
For more information on Magnets
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
The electric field at the point exactly midway between the plates, = 0
3) The electric field, 'E', between plates is given as follows;
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, = E × e
∴ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, ≈ 1.807 × 10⁻⁷ N
Answer:
The frictional torque is
Explanation:
From the question we are told that
The mass attached to one end the string is
The mass attached to the other end of the string is
The radius of the disk is
At equilibrium the tension on the string due to the first mass is mathematically represented as
substituting values
At equilibrium the tension on the string due to the mass is mathematically represented as
The frictional torque that must be exerted is mathematically represented as
substituting values