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SS3: PHYSICS - 1ST TERM

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SS3: PHYSICS – 1ST TERM Electric Field I Theory Questions – Electric Field
Lesson 4, Topic 5
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Theory Questions – Electric Field

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Topic Content:

  • Theory Questions - Electric Field

Theory Questions – Electric Field:

(a) Explain what is meant by: (i) electric field intensity; (ii) electric lines of force.

(b) Two similar but opposite point charges -q and +q each of magnitude 5 

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Question 1

(a) Explain what is meant by: (i) electric field intensity; (ii) electric lines of force.

(i) electric field intensity

Answer:

An electric field is a region in which a force of electrical origin is experienced. Electric Field Intensity (E) at any point is the force per unit charge which the field exerts at that point. The S.I. unit is Newton per coulomb (NC-1) or Volts per metre (Vm-1) . It is a vector quantity and its direction is the same as that of the force or a small positive test charge placed in the field.

⇒ E = \( \frac{F}{q} \)

 

(ii) electric lines of force

Answer:

An electric line of force is an imaginary continuous line or curve drawn in an electric field such that tangent to it at any point gives the direction of the electric force at that point. The direction of a line of force is the path or direction along which a small free positive charge will follow if placed in a field.

The number of lines per unit cross-sectional area is proportional to the magnitude of electric field. The closer the lines, the larger the electric field.

If the curved line above is an electric line of force, then tangents to this line at points A, B, C, and D represent the direction a small positive charge will follow if placed at these points. The direction of motion of the test charge is represented by the arrows.

 

(b) Two similar but opposite point charges -q and +q each of magnitude 5 × 10−8 C are separated by a distance of 8.0 cm in vacuum as shown in the diagram below.

Calculate the magnitude and direction of the resultant electric field intensity E at point P. Draw the lines of force due to this system of charges.

⇒ \(\left( \scriptsize Take \: \normalsize \frac{1}{4 \pi \varepsilon_0 } \scriptsize = 9 \: \times \: 10^9 \: NM^2C^{-2} \right)\)

Solution

Field at P due to -q at P

At P the direction of the field vector is towards the left since the positive test charge at P is attracted to the -q charge.

It is not necessary to use the negative sign for the -q charge because we know the direction of the electric field (to the left). The negative sign shows the path of the field.

E1 = \(  \frac{q_1}{4 \pi \varepsilon_0 r^2 }\)

E1 = \(  \frac{5 \: \times \: 10^{-8} \: \times \: 9 \: \times \: 10^9}{0.05^2 }\)

= 1.8  × 105  NC-1

Field at P due to +q at P

At P the direction of the field vector is towards the left since the positive test charge at P is repelled by the +q charge.

E2 = \(  \frac{q_2}{4 \pi \varepsilon_0 r^2 }\)

E2 = \(  \frac{5 \: \times \: 10^{-8} \: \times \: 9 \: \times \: 10^9}{0.03^2 }\)

= 5  × 105  NC-1

The resultant electric field intensity: E = E1 + E2

E = \( \scriptsize \left( 1.8 \: \times \: 10^5  \: + \: 5 \: \times \: 10^5\right) \: NC^{-1}\)

E = \( \scriptsize 6.8 \: \times \: 10^5 \: NC^{-1}\)

The field is directed towards the left or towards the -q charge.

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