Give the Relation Between Electric Field Intensity and Electric Potential

Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.

Derive the relation between the Electric Field Intensity and the Potential Gradient due to Point Charge at a Point in its Field.

In uniform electric field: Suppose, A and B are two points on a field line in a uniform electric field of intensity E [Fig.], Let d be the distance between the two points and V_A and V_B be the electric potentials at A and B, respectively. If the direction of electric intensity is from A to B, then V_A > V_B and if a free positive charge is placed in this field the charge will move from A to B. Potential difference between A and B = V_A – V_B.

Work done to bring a unit positive charge from B to A = -E ᐧ d [negative sign indicates that intensity and displacement are oppositely directed]
According to the definition of potential difference, this work done is equal to the potential difference between the two points.
∴ V_A – V_B = -E × d
or, E = –\(\frac{V_A-V_B}{d}\) …. (1)
This is the relation between intensity and potential difference in a uniform electric field.

In a non-uniform electric field: Let us consider two points A and B very close to each other in a non-uniform electric field and the distance between the two points = dx [Fig.]. Let the electric field intensity E be directed from A to B.

Since dx is very small, E is practically constant in between A and B. As the direction of intensity is from A to B, V_A > V_B.

Let the potential be V + dV and that at B be V.
The work done to bring a unit positive charge from B to A = -E ᐧ dx.
[negative sign Indicates that intensity and displacement are oppositely directed]
According to the definition of potential difference, this work done is equal to the potential difference between the two points.
∴ (V + dV) – V = -E ᐧ dx
or, E = \(-\frac{d V}{d x}\) ………(2)
This is the relation between Intensity and potential difference in a non-uniform electric field.

Alternative unit of electric field Intensity: The relation E = \(-\frac{d V}{d x}\) suggests that, In CGS system, unit of electric field intensity is statV ᐧ cm^-1 and in SI it is V ᐧ m^-1

Electric potential gradient: In equation (2), \(\frac{d V}{d x}\) is known as potential gradient. Potential gradient is the rate of change of electric potential with respect to displacement. Negative sign indicates the value of potential gradually decreases as we
proceed in the direction of electric field.

Electric Field Intensity and Electric Potential due to a Uniformly Charged Sphere

It is often difficult to calculate electric field intensity and potential due to an irregular shaped charged conductor. But for some regular shaped conductors, intensity and potential can be calculated easily. To calculate intensity and potential due to a uniformly charged sphere at an external point or on its surface, it can be assumed that the whole charge of the sphere is concentrated at its centre.

Suppose a spherical conductor of radius r has q unit of charge [Fig.], For calculation of electric field intensity and potential at P at a distance x from its centre O, it is assumed that the charge q is concentrated at the centre of the sphere.

The electric field intensity at P = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{x^2}\)
[It is assumed that the sphere and the point P are in air]
Electric potential at P = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{x}\), (x > r)

If the point P is situated on the surface of the sphere, then x = r. So electric field intensity on the surface of the sphere = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\); and potential = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\).
It can also be proved that the potential everywhere inside a charged sphere is equal, and the value of this potential is equal to that on its surface, i.e., \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\).
Since the potential inside a sphere is a constant, it is obvious from the relation E = –\(\frac{d V}{d x}\), thai the electric field intensity at every point inside a sphere is zero.

Numerical Examples

Example 1.
A region is specified by the potential function V = 2x² + 3y³ – 5z². Calculate the electric field intensity at a point (2, 4, 5) in this region.
Solution:

Example 2.
Two points A and B are situated at distances 1 m and 2 m from the source of an electrostatic field. The field at a distance x from the source is E = \(\frac{5}{x^2}\). What is the potential difference between A and B?
Solution:

Example 3.
In an electric field, the potential V(x), depending only on the x -coordinate, is given by V(x) = ax – bx³, where a and b are constants. Find out the points on the x -axis where the electric field intensity would vanish. [HS ‘12]
Solution:
E = –\(\frac{d V}{d x}\) = –\(\frac{d}{d x}\)(ax – bx³) = -a + 3bx²
E would vanish, i.e., E = 0, under this condition that,
-a + 3bx² = 0, x² = \(\frac{a}{3 b}\) or, x = ±\(\sqrt{\frac{a}{3 b}}\)
∴ The electric field intensity vanishes at the points
x = \(\sqrt{\frac{a}{3 b}}\) and x = \(-\sqrt{\frac{a}{3 b}}\), on the x-axis