In electrostatics, the concept of Electric field and electric potential plays an important role. Electric field or electric field intensity is the force experienced by a unit positive test charge and is denoted by E. Electric potential is the work done to move unit charge against the electric field or the electric potential difference is the work done by conservative forces to move a unit positive charge and is denoted by V.

Mathematically, the electric field and the potential is given by:

### ⇒ \[E=\frac{F}{q}\]

### ⇒ \[V=\frac{Kq}{r}\]

The relation between electric field and potential is similar to that of the relation between gravitational potential and the field. The relation between Electric field and Potential is generally given by -the electric field is the negative gradient of the electric potential.

## Relation Between Electric Field and Potential:

The relation between electric field intensity and electric potential can be found with a small derivation given below. To establish the relation between Electric field and electric potential we will use basic concepts of electrostatics.

Derivation:

To derive a relation between electric field and potential, consider two equipotential Surfaces separated by a distance dx, let V be the potential on surface 1 and V-dV be the potential on surface 2. Let E be the electric field and the direction of the electric field is perpendicular to the equipotential surfaces.

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Let us consider a unit positive charge +1C near point B, the force experienced by the unit positive charge is given by:

⇒ F= qE ……(1)

Since q= +1C, equation (1) becomes,

⇒F = E………..(2)

Equation (2) is indicating that the magnitude and the direction of the electric field and the force are equal, which means the direction of force is also perpendicular to the equipotential surfaces.

If we move the charge from point B to point A, the work done in bringing the charge from point B to point A is given by,

⇒WBA = F.dx

⇒WBA = F dx Cosθ ……(3)

From equation (2) F = E, Substituting the value of F in equation (3) we get,

⇒WBA = E dx Cosθ …….(4)

Now, the force experienced is acting in the upward direction, but the displacement is in the downward direction, thus the angle between force and displacement is 180°. Therefore, the work done in bringing the point charge from point B to A is given by:

⇒WBA = ─ E dx……….(5)

From the definition of electric potential, we know that the electric potential is the work done in bringing a point charge from one point to another, thus we have:

⇒ WBA = VA- VB

Substituting the corresponding values of the potential at point A and B,

⇒WBA = V- (V-dV)=dV …………..(6)

Equating equation (5) and (6):

⇒ dV = ─ E dx

### ⇒ \[E= -\frac{dV}{dx}\]………………(7)

Therefore, the relationship between field and potential is the electric field due to a point charge is negative potential gradient due to the point charge. Equation (7) is known as the electric field and potential relation.

Equation (7) is the relation between electric field and potential difference in the differential form, the integral form is given by:

We have, change in electric potential over a small displacement dx is:

⇒ dV = ─ E dx

⇒ ∫dV = ─ ∫E.dx

⇒ΔV= V_{A}-V_{B} = ─ ∫E.dx……..(8)

Equation (8) gives the integral form of a relation between the electric field and potential difference.

Case 1:

If the test charge is positive, then from the relationship between the electric field and electric potential, the potential gradient will be more near the charge.

Case 2:

If the test charge is negative, the potential gradient will be more as we move away from the test charge.

Case 3:

For an equipotential surface, the potential at every point on the surface will be the same, thus the potential gradient will be zero. The electric potential will be perpendicular to the electric field lines.

### Examples:

1. Calculate the Electric Potential Due to a Point Charge at a Distance x From it.

Ans:

Given that, a point charge is placed at a distance x from point P(say). We are asked to calculate the potential at point P.

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We know that the electric field due to point charge is given by,

### ⇒ \[E=\frac{kQ}{x^{2}}\]

From the relation between the electric field and the potential we have,

⇒ ∫dV = ─ ∫E.dx

The limit of integration for LHS is 0 to V and for RHS infinity to x.

Substituting the value of E in the above expression,

⇒ \[[V]^{V} 0=\frac{-\int Kq}{x^{2}}.dx\]

⇒ \[V=-Kq\int x ^{2}.dx\]

On simplification we get,

⇒ \[V=\frac{kq}{x}\]

Therefore, by using the relation between the electric field and the potential it is convenient to derive results. The same relation can be derived by using the definition of electric potential.

2. The Electric Potential V at Any Point X, Y, and Z in Space is Given by V=3x2 Volts, then the Electric Field at Any Point (2,1,2) is?

Ans:

Given, the potential at any point x,y, and z,

⇒ V=3x^{2}

We have to find the electric field E at (2,1,2).

We know that,

### ⇒ \[E=-\frac{dV}{dx}\]

### ⇒ \[E=-\frac{d(3x^{2})}{dx}=6x\]

At x=2,

⇒ E=6(2)=12 V/m

Therefore, the electric field E at (2,1,2) is 12V/m.

1. How to Identify the Direction of the Electric Field?

Ans: The direction of the electric field-

The direction of the electric field is always towards the decreasing potential.

If a positive charge is considered, then the direction of the electric field will be away from it, similarly, a negative charge is considered then the direction of the electric field will be towards the negative charge.

The electric field is always perpendicular to the equipotential surfaces.

2. What is the Unit of Electric Field Intensity?

Ans: The electric field intensity can be expressed as,

⇒E=F/q

Thus according to the formula above, the unit of electric field intensity is Newton/Coulomb (N/C).

Or,

⇒E=-dV/dx

Therefore, the unit of electric field intensity is Volts/meter (V/m).