(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.
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)