Two large parallel metal plates carry opposite charges. They are separated by 0.20 m and the p.d. between them is 500V. What is the magnitude of electric field, if it is uniform, in the region between them? Ans: 2500 V/m Solution Separation between the plates (d) = 0.20 m P.d. between them (V) = 500 V Electric field (E) =? We know that, [latex]E=\frac{V}{d}[/latex] = [latex]\frac{500}{0.2} = 2500\ V/m[/latex]
An electron of mass kg charge C is situated in uniform electric field of intensity. Find the time it takes to travel 1 cm from rest. Ans: [latex]3.1\times 10^{-9}[/latex] s Solution Mass of electron (m) = [latex]9.1\times 10^{-31}\ kg[/latex] Charge of electron (q) = [latex]1.6\times 10^{-19}[/latex] C Electric field intensity (E) = [latex]1.2\times 10^4\ Vm^{-1}[/latex] Time (t) =? Distance travelled (s) = 1 cm = 0.01 m First, Acceleration of motion (a) = [latex]\frac{F}{m} = \frac{eE}{m}[/latex] = [latex]\frac{(1.6\times 10^{-19})\times (1.2\times 10^4)}{9.1\times 10^{-31}}[/latex] = [latex]2.109\times 10^{15}[/latex] m/s2 Again, s = [latex]ut+\frac{1}{2}at^2[/latex] Or, [latex]s = 0 + \frac{1}{2}at^2[/latex] Or, [latex]2s = at^2[/latex] Or, [latex]t = \sqrt{\frac{2s}{a}}[/latex] = [latex]\sqrt{\frac{2\times 0.01}{2.109\times 10^{15}}} = 3.079\times 10^{-9}\ s[/latex]
