Electric Field

An electric field is a region where electrical force is being experienced due to the presence of the physical agency charge.

An electric line of force is the path traced by a small positive charge as it moves from electric field.

Electric Field Intensity

Electric field intensity is the force acting on a unit positive charge in an electric field 

E = \( \frac{F}{Q} \)

F = EQ

The unit of electric field intensity (E) is Newton per coulomb (NC^-1) or (VM^-1). It is a vector quantityVector quantities are quantities with magnitude and direction. Examples of vector quantities include displacement, velocity, position, force, and torque. More.

Electric field intensity depends on:

1. Nature of the charge
2. Size of the charge 
3. Distance of the point from the charge

F = EQ

F = \( \frac{1}{4 \pi \varepsilon_0} \frac{Q q}{r^2} \) 

The force between point charge Q and test charge q situated at B is

F = \( \frac{1}{4 \pi \varepsilon_0} \frac{Q q}{r^2} \) 

At B the field intensity is

E = \( \frac{F}{Q} = \frac{1}{4 \pi \varepsilon_0} \frac{Q q}{r^2} \)

E =  \( \frac{Q}{4 \pi \varepsilon_0 r^2} \)

Examples 

1. Calculate the electrostatic field intensity at a point m due to charge q₁. If the positive charge q₂ is of magnitude 10^-8c. Assuming the distance of q₂ from m is 8m

Solution

E =  \( \frac{Q}{4 \pi \varepsilon_0 r^2} \)

=  \( \frac{10^{-8}}{4 \pi \varepsilon_0 \times 8^2} \)

= \( \frac{9 \times 10^9 \; \times \; 10^8}{8^2} \)

=1.41NC^-1

2. A negative charge of magnitude 1.6^-19c is in a uniform electric field intensity 1000VM^-1_. Find

(i) The force experienced by the charge

(ii) Its acceleration if the mass of the charge is 9.1 x 10^-31kg

Solution

(i) E = \( \frac{F}{Q} \)

F = EQ

=1000 x 1.6 x 10^-19

=1.6 x 10^-16 N

(ii) g =\( \frac{F}{M} = \frac {1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \) 

= 1.76 x 10^-14kgms^-1