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Question: \(\text{Joule }{{\left( \text{coulomb} \right)}^{-1}}\) is the same as _ _ _ _ _ _....

Joule (coulomb)1\text{Joule }{{\left( \text{coulomb} \right)}^{-1}} is the same as _ _ _ _ _ _.

Explanation

Solution

Hint: Joule is the unit of energy or work while the coulomb is the unit of electric charge. The work done in moving a charge q in a region of potential V is the product of the charge and the potential (V) present.

Complete step-by-step answer:
As I have mentioned, the work done in moving an electric charge q through a small distance in the region of potential V is given by,W=qV\text{W}=\text{qV}. The work done while moving the charge is stored as electrical energy. We know that the SI unit of work is joules and the SI unit of charge is the coulomb. So if we write the units in the equation, we get
Joules=(Coulomb)×V\text{Joules}=\left( \text{Coulomb} \right)\times \text{V}
V=Joule(Coulomb)1\therefore \text{V}=\text{Joule}{{\left( \text{Coulomb} \right)}^{-1}}
So, Joule (coulomb)1\text{Joule }{{\left( \text{coulomb} \right)}^{-1}} is the same as the electrical potential (V).

Additional Information:
A potential can either be a scalar potential or a vector potential. It can be defined as a field in space from which a lot of physical information can be derived. The gravitational field at a point in space is derived from the gravitational potential at that particular point, and the same is true for the electric field.
The Coulomb potential, the Lennard-jones potential are some examples of potentials associated with a certain force.
The term potential is also used in thermodynamics which is used to signify the thermodynamic potential of a system.

Note: Volt is the SI unit of electric potential. Electric potential is a scalar quantity.
Abvolt and Statvolt are units of electric potential in the centimetre-gram-second system (C.G.S System).
The electric field, which is a vector field, is the gradient of the electric potential. The electric field is a conservative field, which means that the work done in the field is dependent only on the initial and final position of the particle.