PART A)
Electrostatic potential at the position of origin is given by

here we have



now we have


Now work done to move another charge from infinite to origin is given by

here we will have

so there is no work required to move an electron from infinite to origin
PART B)
Initial potential energy of electron




Now we know



now by energy conservation we will have
So here initial total energy is sufficient high to reach the origin
PART C)
It will reach the origin
Answer: An individual living creature is called an organism.
There are many characteristics that living organisms share. All living organisms: respond to their environment.
Answer:
a) B = 1.99 x 10⁻⁴ Tesla
b) B = 0.88 x 10⁻⁴ Tesla
Explanation:
According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:
B = μ₀ I L/4πr²
where,
μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹
I = current = 2 A
L = Length of wire = 40 cm = 0.4 m
a)
r = radius of magnetic field = 2 cm = 0.02 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²
<u>B = 1.99 x 10⁻⁴ Tesla</u>
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b)
r = radius of magnetic field = 3 cm = 0.03 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²
<u>B = 0.88 x 10⁻⁴ Tesla</u>
Answer:
the group of units suggested by the international convention of scientists in 1960 AD to make similarties in meseurment all over the world is called SI units
'Pressure' is (force) / (area).
The only choice with those units is #1 .