(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.
(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 \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect 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 \(20\ \mu\text{F}\).
(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 \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect 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 \(20\ \mu\text{F}\).
