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2000 WAEC Physics Theory (a) Define the following terms: (i) Electric field intensity (ii) Electric potential (b) The diagram...

Physics
WAEC 2000

(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.

 

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Explanation

(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \(\frac{-V}{d}\)

OR E = \(\frac{F}{Q}\). where F is the force,

Q is the charge and E electric field intensity.

(ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field.

 

(b)

 

(c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor

(ii) Factors Effect on capacitor

common area of plates

Distance between plates

Nature of dielectric  

capacitance increases with area

capacitance decreased with increasing distance

capacitance increases with dielectric constant

 

C = \(\frac{KE_oA}{A}\)

(iii) C\(_{2.3}\) = C\(_2\) + C\(_3\) = 20 + 20 = 40\(\mu\) F

\(\frac{1}{C_T} = \frac{1}{C_{2.3}} + \frac{1}{C_1}\)

\(C_T = \frac{\frac{C_{2.3} \times C_1}{C_2.3 + C_1} = \frac{40 \times 40}{40 \times 40}\)

= 20\(\mu F\)


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