Question

In the Van de Graaff generator shown in Figure 16.78, a rubber belt carries electrons up through a small hole in a large hollow spherical metal shell. The electrons come off the upper part of the belt and drift through a wire to the outer surface of the metal shell, so that the metal shell acquires a sizable negative charge, approximately uniformly distributed over the sphere. At a time when the sphere has acquired a sizable charge -Q, approximately how much work must be done by the motor to move one more electron from the base (a distance h below the sphere) to the upper pulley (located a distance R/2from the center of the hollow sphere)? Explain your work, and state explicitly what approximations you had to make.

Short answer

The work must be done by the motor to move one more electron from the base to the upper pulley \(\frac{{Q{q_e}}}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\).

Step-by-step solution

Write the given data from the question.

The distance below the sphere is\(h\).

The radius of the sphere is\(R\).

The charge on the sphere is\( - Q\).

Determine formulas to calculate the work must be done by the motor to move one more electron from the base to the upper pulley.

The expression to calculate the electrical potential at the distance \(r\) is given as follows.

\(V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r}\)

The expression to calculate the work done is given as follows.

\(W = {q_e}\Delta V\)

Here,\(\Delta V\)is the change in the potential.

Calculate the work must be done by the motor to move one more electron from the base to the upper pulley.

Calculate the electric potential at the centre of the sphere.

Substitute \(R\) for \(r\) and \(Q\) for \(q\) into equation (i).

\({V_{centre}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R}\)

The total distance from the centre to the bottom is\(R + h\).

Calculate the electric potential at the centre of the sphere.

Substitute \(R + h\) for \(r\) and \(Q\) for \(q\) into equation (i).

\({V_{bottom}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{R + h}}\)

Calculate the change in the electric potential.

\(\Delta V = {V_{centre}} - {V_{bottom}}\)

Substitute \(\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R}\) for \({V_{centre}}\) and \(\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{R + h}}\) for \({V_{bottom}}\) into above equation.

\(\begin{array}{l}\Delta V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{R + h}}\\\Delta V = \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\end{array}\)

Calculate the work done.

Substitute \(\frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\) for \(\Delta V\) into equation (ii).

\(\begin{array}{l}W = {q_e}\frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\\W = \frac{{Q{q_e}}}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\end{array}\)

Hence the work must be done by the motor to move one more electron from the base to the upper pulley\(\frac{{Q{q_e}}}{{4\pi {\varepsilon _0}}}\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\).