Electric Field: Electric field is that space in which a charged particle experiences a force. Electric field strength or Electric Intensity at a point in an electric field is the force per unit charge acting on a small test charge kept at that point. Mathematically

$ \displaystyle \vec{E} = \frac{\vec{F}}{q} $ ; where the test charge is always positive

Electric intensity at a distance r from a point charge Q is given as

$ \displaystyle E = k \frac{Q}{r^2} = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$

If electric intensity is same at all points in the region then field is said to be uniform. Equispaced parallel lines represent uniform electric field. Arrow on the lines indicates the direction of the electric field.

The electric field at a point due to several discrete charges distributed in space is the vector sum of the fields due to individual charges at the point (principle of superposition)

$ \displaystyle \vec{E} = \vec{E_{1}} + \vec{E_{2}} +\vec{E_{3}} ……. + \vec{E_{n}}$

In case of a continuous charge distribution we can use the technique of integration to find the resultant electric field at a point. The electric field of a continuous charge distribution at some point is given by

$ \displaystyle \vec{E} = \frac{1}{4\pi \epsilon_0}\int\frac{dq}{r^2}\hat{r} $

where dq is the charge on one element of the charge distribution and r is the distance from the element to the point under consideration and r ˆ is the unit vector directed from the position of elemental charge towards the point where electric field is to be found out.

Before taking integration, it is important to be sure that all the components of the parts to be integrated have the electric field at that point in the same direction.

Illustration : Find the electric field on the axis of a charged disc.

Solution: Consider a disc of uniform surface charge density ‘ σ ‘ . Let us calculate the electric field due to a ring of charge situated at a distance r , from the centre and having a width, dr.

Here we are using the expression for electric field intensity for a charged ring of radius r at a point on the axis at a distance x from the centre,

$ \displaystyle E = \frac{k x q}{(x^2 + r^2)^{3/2}} $ ; directed along the axis outwards from the centre.

$ \displaystyle dE = \frac{k x dq}{(x^2 + r^2)^{3/2}} $ ; directed along the line OP

Now, the area of the ring, dS = 2πr. dr , dq = σ (2πrdr ),

Thus ,

$ \displaystyle E_p = \int dE = k x \int_{0}^{R}\frac{\sigma 2\pi rdr }{(x^2 + r^2)^{3/2}}$

$ \displaystyle E = \frac{\sigma}{2 \epsilon_0}[1 – \frac{x}{(x^2 + R^2)^{1/2}}] $

$ \displaystyle E = \frac{\sigma}{2 \epsilon_0}[1 – cos\theta] $ ; where θ = semi vertical angle subtend by the disc at P.

Also, as R → ∞

$ \displaystyle E = \frac{\sigma}{2\epsilon_0} $ ; which is the electric field in front of an infinite plane sheet of charge.

Illustration : Find the electric field intensity at a point P which is at a distance R (point lying on the perpendicular drawn to the wire at one of its end) from a semi-infinite uniformly charged wire. (Linear charge density = λ )

Solution : Field at point P is due to an elemental charge dq ( = λ(dl) )

$ \displaystyle dE = k\frac{dq}{(R^2 + l^2)} $

The component along x-axis

$ \displaystyle dE_x = k\frac{dq cos\theta}{(R^2 + l^2)} $

$ \displaystyle E_x = \int_{0}^{\infty}\frac{1}{4\pi \epsilon_0}\frac{\lambda dl R}{(R^2 + l^2)^{3/2}} $

$ \displaystyle E_x = \frac{\lambda}{4 \pi \epsilon_0 R} $

$ \displaystyle E_y = \frac{\lambda}{4 \pi \epsilon_0 R} $

$ \displaystyle E = \sqrt{E_x^2 +E_y^2} $

$ \displaystyle E = \frac{\lambda}{2\sqrt2 \pi \epsilon_0 R} $

$ \displaystyle tan\theta = \frac{E_y}{E_x} $

⇒ θ = 45°