three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
Length of the pendulum (l) = 2 m
Acceleration due to gravity = g = 9.8 m/s^2
For small amplitude, the pendulum will undergo simple harmonic motion.
Hence, the time period of the pendulum for small amplitude =
Now, plug the values of l and g
T =
T = 2 × 3.14 × 0.451
T = 2.83 seconds
Hence, the time period of the pendulum for small amplitude = 2.83 s
1. The problem statement, all variables and given/known data A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. 2. Relevant equations When a capacitor is charged, the electric field E, and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: ∮B⃗ ⋅dl⃗ =μ0(I+ϵ0dΦdt). You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero.
I think its d . i hope this helps
To produce an electric current, three things are needed: a supply of electric charges, also called electrons, which are free to flow, some form of push to move the charges through the circuit and a pathway to carry the charges. The pathway to carry the charges is usually a copper wire.