A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 0.9 µF capacitor then drops to 2 V. What is the capacitance of the second capacitor?

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

for this one it would be d bump/forearm pass

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: