Question

(a) Define electric field and electric field intensity at a point.

(b) Find the electric field intensity

(i) due to a group of point charges and

(ii) due to a general charge distribution.

Answer 1

(a) Electric field. The electric field due to a point charge or a system of charges is the space around the charge within which the force of attraction/ repulsion due to the charge can be experienced. Theoretically, the electric field due to a charge extends upto infinity. But practically, it is confined only to some distance from the charge.

Electric field intensity. Electric, field intensity (\(\vec E\))  at any point in the electric field is equal to the force acting over a unit positive charge on a vanishingly small charge placed at that point.

If \(\vec F\)  is the coulomb force acting on a small positive test charge q0, then by definition,

\(\vec E= Lt_{q_0}\rightarrow\frac{\vec F}{q_0} \)

The S.I. unit of is newton/coulomb. It is a vector quantity and the direction of \(\vec E\)  is the direction in which the test charge will move if free to do so.

Field intensity due to a point charge. Suppose we have to calculate electric field intensity at any point P to a point charge q at O, where OP = r. Imagine a positive test charge q0 at P. According to Coulomb’s law, force at P is

\(\vec F= \frac{1}{4\piɛ_o}\frac{qq_0}{r^2}\hat r\) where \(\hat r\) is unit vector directed from q towards q₀ as shown in Fig. below.

\(\hat r\)  is the unit vector in the direction of force.
If q is + ve, the field intensity is directed radially outwards along OP. If q is negative, the field intensity is directed towards q, along PO. 

(b) (i) Electric field intensity due to a group of charges. The electric field intensity at any point due to a group of point charges is equal to the vector sum of the electric field intensities due to given charges at the same point. For this, we first calculate the electric field intensity at the given point due to individual charges and then add them vectorially. Let us consider a system of n charges q₁, q₂, ................. q_n with position vectors \(\vec r_1,\vec r_2,.....\vec r_n\)  respectively relative to some origin O. Let a unit charge be placed at origin O as shown in Fig., then

Electric field due to q₁ at \(\vec r_1\) is

\(\vec E_1 = \frac{1}{t.\piɛ_o} \frac{q_1}{r^2_1}\hat r\)

Electric field due to q₂ at \(\vec r_2\) is

\(\vec E_2 = \frac{1}{4\piɛ_o} \frac{q_2}{r^2_2}\hat r_2\)

Similarly

\(\vec E_n = \frac{1}{4\piɛ_o} \frac{q_n}{r^2_n}\hat r_n\)

∴ Total electric field at 0 due to the system of charges is given by

Here ri is the distance of the point P from the ith charge q_i and \(\hat r\)is unit vector directed from qi to the point P.

(ii) General expression for the electric field at a point. The general expression for the force on a charge q₀ at \(\hat r\)  due to the most general (static) source distribution is given by

Now electric field \(\vec E\) at a point is given by

\(\vec E = Lt_{q_0} \rightarrow0\frac{\vec F}{q_0}\)......................(i)

Putting the value of \(\vec F\), we get