(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)(i) 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 SI unit is \(\mathrm{N\,C^{-1}}\) (or \(\mathrm{V\,m^{-1}}\)), and its direction is the direction of the force on a positive test charge.

(ii) Electric lines of force are imaginary lines used to show an electric field. The tangent to a field line at any point gives the direction of the field. They emerge from positive charges and terminate on negative charges; closer spacing indicates a stronger field, and field lines never intersect.

(b) Point \(P\) lies between the charges, \(0.050\,\mathrm{m}\) from \(-q\) and \(0.030\,\mathrm{m}\) from \(+q\). Taking \(k=\dfrac{1}{4\pi\varepsilon_0}=9.0\times10^9\,\mathrm{N\,m^2\,C^{-2}}\):

Field at \(P\) due to \(-q\):

\[E_{-}=\frac{kq}{(0.050)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.050)^2}=1.80\times10^5\,\mathrm{N\,C^{-1}}\]

Field at \(P\) due to \(+q\):

\[E_{+}=\frac{kq}{(0.030)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.030)^2}=5.00\times10^5\,\mathrm{N\,C^{-1}}\]

At \(P\), both fields are directed from \(+q\) towards \(-q\). Therefore,

\[E=E_{+}+E_{-}=5.00\times10^5+1.80\times10^5=6.80\times10^5\,\mathrm{N\,C^{-1}}\]

Resultant field intensity at \(P\) = \(6.8\times10^5\,\mathrm{N\,C^{-1}}\), directed towards the negative charge (from \(+q\) to \(-q\)).

The electric lines of force for the two unlike point charges are shown below.

(c)

Given: \(f=25\,\mathrm{Hz}\), \(C=10\,\mu\mathrm{F}=10\times10^{-6}\,\mathrm{F}\), \(R=1000\,\Omega\), and \(V=90\,\mathrm{V_{rms}}\).

(i) Capacitive reactance

\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi(25)(10\times10^{-6})}=636.6\,\Omega\]

(ii) Impedance

\[Z=\sqrt{R^2+X_C^2}=\sqrt{1000^2+636.6^2}=1185.4\,\Omega\]

(iii) Current through the circuit

\[I=\frac{V}{Z}=\frac{90}{1185.4}=7.59\times10^{-2}\,\mathrm{A}=0.0759\,\mathrm{A_{rms}}\]

(iv) Voltage across the capacitor

\[V_C=IX_C=(0.0759)(636.6)=48.3\,\mathrm{V_{rms}}\]

(v) Average power used in the circuit

Only the resistor dissipates average power:

\[P=I^2R=(0.0759)^2(1000)=5.76\,\mathrm{W}\]