Question

On moving a charge of $20 \mathrm{C}$ by $2 \mathrm{cm}, 2$ J of work is done, then the potential difference between the points?
(A) $0.1 \mathrm{V}$
(B) $8 \mathrm{V}$
(C) $2 \mathrm{V}$
(D) $0.5 \mathrm{V}$

Answer 1

Hint: We know that electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. The space around an electric charge in which its influence can be felt is known as the electric field. The electric field Intensity at a point is the force experienced by a unit positive charge placed at that point. Electric Field Intensity is a vector quantity.

Complete step by step answer
We know that voltage, also sometimes called potential difference or electromotive force (EMF), refers to the amount of potential energy the electrons have in an object or circuit. In some ways, you can think of this as the amount of "push" the electrons are making to try to get towards a positive charge. Therefore, a system consisting of a negative and a positive point-like charge has a negative potential energy. A negative potential energy means that work must be done against the electric field in moving the charges apart.
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honour of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb. Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage.
Voltage is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V). Voltmeters are used to measure the potential difference between two points. There is a misconception about potential and voltage. Many of us think that both are the same. But voltage is not exactly potential; it is the measure of the electric potential difference between two points.

The potential difference between two points in an electric field is, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\dfrac{\mathrm{W}}{\mathrm{q}_{0}}$
where $\mathrm{W}$ is work done by moving charge $\mathrm{q}_{0}$ from point $\mathrm{A}$ to $\mathrm{B}$. $\mathrm{So}, \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\dfrac{2}{20}=0.1 \mathrm{V}$

So, the correct answer is option A.

Note: It can be said that electric fields (e-fields) are an important tool in understanding how electricity begins and continues to flow. Electric fields describe the pulling or pushing force in a space between charges. The electric fields of single charges. A negative charge has an inward electric field because it attracts positive charges. Electric field is not negative. It is a vector and thus has negative and positive directions. An electron being negatively charged experiences a force against the direction of the field. For a positive charge, the force is along the field. There are two types of electric fields: static (or electrostatic) fields and dynamic (or time-varying) fields. Electric fields have a definite magnitude and specific direction.