The strength of the electric field is determined by Solution: As the question states, the area of the electrode is given, and the charge, q, equals 50 nC. To compute the electric field strength, we need to first ascertain the surface charge density which is defined as... Thus, the electric field strength above the center of the electrode can be calculated as:
<span>You are presented with a circuit that includes a 6.0-v battery, a 4.0-ohm resistor, a 0.60 microfarad capacitor, an ammeter, and a switch all connected in series. Your task is to determine the current reading once the switch is closed. Ohm's law should be used, which states V = IR where V signifies voltage, I indicates current, and R represents resistance.</span>
V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A
Upon closing the switch, the cathode side plate starts accumulating electrons if it was previously empty. As this process continues, the current diminishes. Eventually, when the capacitor reaches its maximum electron retention, the current will cease. An increased capacitance means a greater capacity for electron storage.
Answer:
Explanation:
To begin with, we must determine the pressure acting on the sphere, which is calculated using:
where
denotes the atmospheric pressure
represents the density of the water
signifies the acceleration due to gravity
indicates the depth
By substituting these values,
The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m
Thus, the sphere's total surface area can be expressed as
Consequently, the inward force acting on the sphere equals
