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Lesson 4, Topic 2
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# Electric Field

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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 quantity.

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

error: