(a) Define the following terms: (i) Electric field intensity (ii) Electric potential (b) The diagram below illustrates two collinear electric charges of mag...

Assessment: WAEC SSCE - Physics - 2000 (Essay) Subject: Physics

Question 1 Report

(a) Define the following terms:

(i) Electric field intensity

(ii) Electric potential

(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.


Copy the diagram and use arrows to indicate, from the point P, the direction of the;

(i) electric force F\(_1\) due to + Q.

(ii) electric force F\(_2\) due to -Q.

(iii) electric field intensity E.

 

(c) What is meant by dielectric substance?

(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.

(iii)

The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.

 

Answer Details

(a)(i) Electric field intensity at a point is the electric force experienced per unit positive charge placed at that point, \( E = \dfrac{F}{Q} \), measured in \( \text{N C}^{-1} \) (or \( \text{V m}^{-1} \)). It is a vector quantity.

(a)(ii) Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, \( V = \dfrac{W}{Q} \), measured in volts (\( \text{J C}^{-1} \)). It is a scalar quantity.

(b) The charges \(+Q\) and \(-Q\) are collinear, with the rest charge at \(P\) lying on the same line midway between them. Taking the rest charge as positive:

  • (i) Force \(F_1\) due to \(+Q\): like charges repel, so the arrow points away from \(+Q\), that is from \(P\) directed towards \(-Q\).
  • (ii) Force \(F_2\) due to \(-Q\): unlike charges attract, so the arrow also points from \(P\) directed towards \(-Q\).
  • (iii) Resultant field intensity \(E\): both forces act in the same direction along the line, so \(E\) is directed from the \(+Q\) side towards the \(-Q\) side, with magnitude equal to the sum of the two contributions.

(c)(i) A dielectric substance is an insulating (non-conducting) material placed between the plates of a capacitor; it becomes polarised in the field and increases the capacitance.

(c)(ii) Factors determining the capacitance of a parallel-plate capacitor \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):

FactorEffect on capacitance
Common (overlap) area of the plates, \(A\)Capacitance increases as the area increases (directly proportional).
Distance between the plates, \(d\)Capacitance decreases as the separation increases (inversely proportional).
Nature of the dielectric (dielectric constant \(K\))Capacitance increases as the dielectric constant increases.

(c)(iii) Effective capacitance of the section. \(C_2\) and \(C_3\) are in parallel:

\[ C_{2,3} = C_2 + C_3 = 20 + 20 = 40\ \mu\text{F}. \]

This combination is in series with \(C_1 = 40\ \mu\text{F}\):

\[ \frac{1}{C_T} = \frac{1}{C_{2,3}} + \frac{1}{C_1} \quad\Rightarrow\quad C_T = \frac{C_{2,3}\times C_1}{C_{2,3}+C_1} = \frac{40\times 40}{40+40} = 20\ \mu\text{F}. \]

The effective capacitance of the section is \(20\ \mu\text{F}\).

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