Explanation:
it is given that, the linear charge density of a charge, 
Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :
..........(1)
The linear charge density is given by :


Integrating equation (1) from x = x₀ to x = infinity



Hence, this is the required solution.
Answer:
100 V
Explanation:
Hi there!
Ohm's law states that
where V is the voltage, I is the current and R is the resistance.
Plug the given information into Ohm's law (R=50, I=A)

Therefore, the voltage across this current is 100 V.
I hope this helps!
A group of protons and neutrons surrounded by electrons
Answer:
To write the general formula for an acid, we fix one atom which is hydrogen because this atom is common to all the acids. General formula for acid is written by HX. where H represents Hydrogen atom.
Explanation:
Answer:
Mass and height
Explanation:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 
Which is represented as;

stands for gravitational potantial energy,
m stands for mass of object,
g is the gravitational constant and
h is the height.
Here we see that mass of object and height is directly proportional to the gravitational potential energy.
That means increasing in mass and height will result in increasing gravitational potential energy.