Three very large charged metal plates are arranged as shown in Figure 16.77. The radius of each plate is\(4\;m\), and each plate is \(w = 0.05\;mm\) thick. The separation \({d_1}\)is\(6\;mm\), and the separation \({d_2}\)is\(2\;mm\). Each plate has a tiny hole in it, so it is possible for a small charged particle to pass through all the plates.

You are able to adjust the apparatus by varying the electric field in the region between location \(D\)and location\(F\). You need to adjust this setting so that a fast-moving electron moving to the right, entering at location\(A\), will have lost exactly \(5.2 \times 1{0^{ - 18}}\;J\)of kinetic energy by the time it reaches location\(G\). Using a voltmeter, you find that the potential difference\({V_C} - {V_B} = - 16\;V\). Based on this measurement, you adjust the electric field between \(D\) and \(F\)to the appropriate value. (a) Consider the system of (electron \( + \)plates). Neglecting the small amount of work done by the gravitational force on the electron, during this process (electron going from \(A\)to\(G\)), what is\(\Delta K + \Delta U?\)? (b) What is the change in potential energy for the system during this process? (c) What is\({V_G} - {V_A}\)? (d) What is\({V_F} - {V_D}\)? (e) What is the electric field (magnitude and direction) in the region between locations \(D\) and\(F\)?

The radius of the plate,\(r = 4\;{\rm{m}}\)

The thickness of the plate,\(w = 0.05\;{\rm{mm}}\)

The separation distance,\({d_1} = 6\;{\rm{mm}}\)

The separation distance,\({d_2} = 2\;{\rm{mm}}\)

Lost kinetic energy, \(\Delta {U_{GA}} = 5.2 \times {10^{ - 18}}\;{\rm{J}}\)

The potential difference between \(B\) and\(C\),\({V_C} - {V_B} = - 16\;{\rm{m}}\)

The expression to calculate the potential difference between the two points is given as follows.

\(\Delta V = \frac{{\Delta U}}{q}\)……. (i)

Here,\(\Delta U\)is the change in the electrical potential energy and\(q\)is the charge.

The expression to calculate the electric field between the two plates is given as follows.

\(E = \frac{{\Delta V}}{d}\) …… (ii)

Here,\(d\)is the separation distance between the plates.

The expression to calculate the capacitance between the parallel plates is given as follows.

\(C = \frac{{A{\varepsilon _0}}}{d}\) …… (iv)

Here,\(A\)is the area of the plate and\({\varepsilon _0}\)is the permittivity of the free space.

(e)

Calculate the electric field between \(D\) and \(F\).

Substitute \({V_F} - {V_D}\) for\(\Delta V\), and \({d_2}\) for \(d\) into equation (ii).

\(E = \frac{{{V_F} - {V_D}}}{{{d_2}}}\)

Substitute -7.2 x 10^-13 V for V_F- V_Dand 2 mm for d₂ into above equation.

E = (-7.2 x 10^-13 ) / (2 x 10³)

E =3.6 x 10^-10 V/m.

Hence the electric field between Dand F -3.6 x 10^-10 V/m.