(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 magn...
(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 \(5 \times 10^{-8} C\) are seperated by a distance of 8.0cm in vacuum as shown in the diagram below.
Calculate the magnitude and direction of the resultant electric field intensity E at the point P. Draw the lines of force due to this system of charges. [Take \(\frac{1}{4 \pi \varepsilon _{0}}\)]
(c)
Calculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.
(a)(i) Electric field intensity
Electric field intensity, \(E\), at a point is the force experienced per unit positive test charge placed at that point:
\[E=\frac{F}{q}\]
It is a vector quantity. Its direction is the direction of the force on a positive test charge. Its SI unit is \(\text{N C}^{-1}\), equivalently \(\text{V m}^{-1}\).
(a)(ii) Electric lines of force
Electric lines of force are imaginary lines used to show an electric field pattern. The tangent to a line at any point gives the direction of the electric field at that point. Lines leave positive charges and enter negative charges. They never cross, because the field cannot have two different directions at one point. Where the lines are closer together, the electric field is stronger.
(b) Resultant electric field intensity at \(P\)
The distances shown are \(0.03\ \text{m}\) from \(+q\) and \(0.05\ \text{m}\) from \(-q\). Since these add to \(0.08\ \text{m}\), point \(P\) lies between the charges.
The field due to a positive charge points away from it. At \(P\), this is towards the negative charge. The field due to a negative charge points towards it, which is also towards the negative charge. Therefore, the two fields act in the same direction and must be added.
Result: \(\boxed{6.8\times10^5\ \text{N C}^{-1}}\), directed from \(+q\) towards \(-q\).
The supplied reference answer subtracts the two field magnitudes. That would only be appropriate if the fields were in opposite directions. Here \(P\) is between opposite charges, so both fields point towards \(-q\).
(c) Series \(R\)-\(C\) circuit
Using the circuit values shown in the supplied reference material: \(R=1000\ \Omega\), \(C=10\times10^{-6}\ \text{F}\), \(f=25\ \text{Hz}\), and \(V=90\ \text{V}_{\rm rms}\).
Only the resistor dissipates average power. An ideal capacitor stores and returns energy, so its average power is zero.
\[
P=I^2R
=(0.0759)^2(1000)
=5.76\ \text{W}
\]
The value \(I^2Z=6.83\) is not the average power in watts; it is associated with apparent power, \(VI\), measured in volt-amperes. For average power in an \(R\)-\(C\) circuit, use \(I^2R\) or \(VI\cos\phi\).
Examination reminder: decide whether electric-field vectors point in the same or opposite directions before adding their magnitudes. In an a.c. circuit containing a capacitor, use \(R\), not \(Z\), when calculating real average power.
Electric field intensity, \(E\), at a point is the force experienced per unit positive test charge placed at that point:
\[E=\frac{F}{q}\]
It is a vector quantity. Its direction is the direction of the force on a positive test charge. Its SI unit is \(\text{N C}^{-1}\), equivalently \(\text{V m}^{-1}\).
(a)(ii) Electric lines of force
Electric lines of force are imaginary lines used to show an electric field pattern. The tangent to a line at any point gives the direction of the electric field at that point. Lines leave positive charges and enter negative charges. They never cross, because the field cannot have two different directions at one point. Where the lines are closer together, the electric field is stronger.
(b) Resultant electric field intensity at \(P\)
The distances shown are \(0.03\ \text{m}\) from \(+q\) and \(0.05\ \text{m}\) from \(-q\). Since these add to \(0.08\ \text{m}\), point \(P\) lies between the charges.
The field due to a positive charge points away from it. At \(P\), this is towards the negative charge. The field due to a negative charge points towards it, which is also towards the negative charge. Therefore, the two fields act in the same direction and must be added.
Result: \(\boxed{6.8\times10^5\ \text{N C}^{-1}}\), directed from \(+q\) towards \(-q\).
The supplied reference answer subtracts the two field magnitudes. That would only be appropriate if the fields were in opposite directions. Here \(P\) is between opposite charges, so both fields point towards \(-q\).
(c) Series \(R\)-\(C\) circuit
Using the circuit values shown in the supplied reference material: \(R=1000\ \Omega\), \(C=10\times10^{-6}\ \text{F}\), \(f=25\ \text{Hz}\), and \(V=90\ \text{V}_{\rm rms}\).
Only the resistor dissipates average power. An ideal capacitor stores and returns energy, so its average power is zero.
\[
P=I^2R
=(0.0759)^2(1000)
=5.76\ \text{W}
\]
The value \(I^2Z=6.83\) is not the average power in watts; it is associated with apparent power, \(VI\), measured in volt-amperes. For average power in an \(R\)-\(C\) circuit, use \(I^2R\) or \(VI\cos\phi\).
Examination reminder: decide whether electric-field vectors point in the same or opposite directions before adding their magnitudes. In an a.c. circuit containing a capacitor, use \(R\), not \(Z\), when calculating real average power.
