<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:
a
The value at a point inside is Zero
b
The electric field is 
Explanation:
We know from the problem that
The charge magnitude is 
The radius of the spherical ball is 
According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero
On the outside, the electric field around the spherical ball is mathematically expressed as

Here a denotes a point outside the spherical ball with its value of 
and k represents Coulomb's constant, valued at

=> 
=> 
Answer:
x = v₀ cos θ t, y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.
Now, we will apply trigonometry to determine the initial velocity components.
sin θ =
/ v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = v₀ cos θ
Next, let's formulate the equations of motion.
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ +
t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components