Question

Assume that an electric field ${{\vec E = 30}}{{{x}}^{{2}}}{{\hat i}}$ exists in space. Then the potential difference ${V_A} - {V_O}$ , where ${V_O}$ is the potential at the origin and ${V_A}$ the potential at ${{x = 2m}}$ is
(A) -80V
(B) 80V
(C) 120V
(D) -120V

Answer 1

Hint : Integration of the electric field with respect to the distance of two points gives the potential difference of the two points. The limits of integration are set at ${{x = 2m}}$ and at the origin, that is, zero.

Formula Used: The formulae used in the solution are given here.
$V = Ed$ . In this equation, $V$ is the potential difference in volts, $E$ is the electric field strength (in Newton per coulomb), and $d$ is the distance between the two points (in meters).

Complete step by step answer
In a uniform electric field, the equation to calculate the electric potential difference is super easy: $V = Ed$ . In this equation, $V$ is the potential difference in volts, $E$ is the electric field strength (in Newton per coulomb), and $d$ is the distance between the two points (in meters).
It has been given that an electric field ${{\vec E = 30}}{{{x}}^{{2}}}{{\hat i}}$ exists in space.
The potential difference between two points ${V_O}$ and ${V_A}$ , ${V_O}$ is the potential at the origin and ${V_A}$ the potential at a point 2m has to be found out.
Potential difference between any two points in an electric field is given by,
$dV = - \int {E \cdot dx}$ where $E$ is the electric field and $x$ is the distance.
Since, electric field ${{\vec E = 30}}{{{x}}^{{2}}}{{\hat i}}$ ,
$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^{20} {30{x^2}dx} }$
${V_A} - {V_O} = \left[ {30\dfrac{{{x^3}}}{3}} \right]_0^2 = \left[ {10{x^3}} \right]_0^2$
Substituting the values of $x$ in the given equation,
$- 10\left[ {{2^3} - {0^3}} \right] = - 80V$
Hence the potential difference ${V_A} - {V_O}$ , where ${V_O}$ is the potential at the origin and ${V_A}$ the potential at ${{x = 2m}}$ is -80V.
The correct answer is Option A.

Note
The electric potential difference is the difference in electric potential between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is ${{\Delta V = }}\dfrac{{{{Work}}}}{{{{Charge}}}}$ .
Electric potential difference, also known as voltage, is the external work needed to bring a charge from one location to another location in an electric field. Electric potential difference is the change of potential energy experienced by a test charge that has a value of +1.
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. If the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations.