The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
<h3>
What is the energy in a capacitor?</h3>
The energy stored in a capacitor is an electrostatic potential energy.
It is related to the charge(Q) and voltage (V) between the capacitor plates.
It is represented as 'U'.
<h3>
How to determine the potential difference</h3>
Formula:
Potential difference, V is the ratio of the charge to the capacitance of a capacitor.
It is calculated using:
V = Q ÷ C
Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9
Substitute the values into the equation
Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts
<h3>
How to determine the energy stored</h3>
Formula:
Energy, U = 1 ÷ 2 (QV)
Where Q= charge and V = potential difference across the capacitor
Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)
= 0.5 × 25 × 10∧-1
= 0.5 × 2.5
= 1. 25 Joules
Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
Learn more about capacitance here:
brainly.com/question/14883923
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Explanation:
the average velocity of the car is 15 m/s example I have this on a test
Answer:
Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.
Answer:
a)n= 3.125 x 
electrons.
b)J= 1.515 x 
A/m²
c)
=1.114 x 
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x 
m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x 
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x 
C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
= 
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
D. 5.0A because this is right and will lead to the right answer okay you got this girl letssssss goooo googoggo Gogol
