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Question: If \({V_0}\) be the potential at origin in an electric field \(\overrightarrow E = {E_x}\widehat i +...

If V0{V_0} be the potential at origin in an electric field E→=Exi^+Eyj^\overrightarrow E = {E_x}\widehat i + {E_y}\widehat j, then the potential at point P(x,y)P\left( {x,y} \right) is
A) V0+xEx+yEy{V_0} + x{E_x} + y{E_y}
B) V0+xEx−yEy{V_0} + x{E_x} - y{E_y}
C) V0−xEx−yEy{V_0} - x{E_x} - y{E_y}
D) x2+y2Ex2+Ey2−V0\sqrt {{x^2} + {y^2}} \sqrt {E_x^2 + E_y^2} - {V_0}

Explanation

Solution

The electric potential at a point in the field is defined as the work done in bringing a unit positive charge from infinity to the point in the electric field. The relationship between this quantity of electric potential and the electric field should be established to solve this problem.

Complete step by step solution:
The electric potential is defined as the work done in bringing a unit positive charge from infinity to a point in the electric field.
Electric potential,
V=WqV = \dfrac{W}{q}
where W = work done and q = charge.
Work done, here, refers to the force acted on moving the charge across a distance d in the electric field. So,
W=FdW = Fd
Substituting we have –
⇒V=Fqd\Rightarrow V = \dfrac{F}{q}d
The electric field is a quantity which is defined by the force acting on the charge. It is given as–
⇒E=Fq\Rightarrow E = \dfrac{F}{q}
Thus, we have –
⇒V=Ed\Rightarrow V = Ed
The electric potential at a point distant d, is equal to the product of electric field at the point and the distance d.
Consider a positive charge at origin with potential at origin as V0{V_0}.

Let P(x,y)P\left( {x,y} \right) be a point in the x-y such that the unit vector joining the origin
with P is given by –
⇒d→=xi^+yj^\Rightarrow \overrightarrow d = x\widehat i + y\widehat j
The electric field at the point P(x,y)P\left( {x,y} \right) is equal to,
⇒E→=Exi^+Eyj^\Rightarrow \overrightarrow E = {E_x}\widehat i + {E_y}\widehat j
The electric potential at the point P due to the charge at origin is given by –
⇒VP=Ed\Rightarrow {V_P} = Ed
Since E and d are in vectors, in vector notation, we have the electric potential as the dot-product of the vectors E and d as shown:
⇒VP=E→⋅d→\Rightarrow {V_P} = \overrightarrow E \cdot \overrightarrow d
⇒VP=(Exi^+Eyj^)⋅(xi^+yj^)\Rightarrow {V_P} = \left( {{E_x}\widehat i + {E_y}\widehat j} \right) \cdot \left( {x\widehat i + y\widehat j} \right)
⇒VP=xEx+yEy\Rightarrow {V_P} = x{E_x} + y{E_y}
The absolute potential at the point P is given as VP=xEx+yEy{V_P} = x{E_x} + y{E_y} . However, the potential at the origin is given as V0{V_0}.
So, the potential at P is equal to the potential difference between the potential at O and potential at P.
⇒V=V0−VP\Rightarrow V = {V_0} - {V_P}
Substituting, we get –
⇒V=V0−VP\Rightarrow V = {V_0} - {V_P}
⇒V=V0−(xEx+yEy)\Rightarrow V = {V_0} - \left( {x{E_x} + y{E_y}} \right)
∴V=V0−xEx−yEy\therefore V = {V_0} - x{E_x} - y{E_y}

Hence, the correct option is Option C.

Note: The unit of electric field through the definition of the electric field is newton per coulomb (N/C). However, another unit of electric field, volt per metre (V/m) is derived from this expression. Both of the units are used to denote the electric field while the latter is called a derived unit and the former is a fundamental unit.