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.