Solved Problem on Work and Electric Potential
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A charge Q = −3 μC is fixed at a point O of the space. Points A, B, and C are at a distance, respectively, 1.0 m, 3.0 m, and 6.0 m from O. The charge is placed in the vacuum, where \( k_{0}=9\times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}} \).
a) Calculate and draw the electric field vector in B.
b) What is the electrostatic potential in B?
c) What is the potential energy of a particle of q = −5 nC placed in C? Consider the potential energy equal to zero in infinity;
d) What is the work of an external agent, necessary to bring the q particle from infinity to point C?
e) What is the work of the electric force in this displacement?
f) What is the work of an external agent when q is displaced from C to the A?
g) What is the work of the electric force in this displacement?



Problem Data:
  • Charge Q:    Q = −3 μC = −3 ×10−6 C;
  • Charge q:    q = −5 nC = −3 ×10−9 C;
  • Distance \( \overline{OA} \):    rA = 1.0 m;
  • Distance \( \overline{OB} \):    rB = 3.0 m;
  • Distance \( \overline{OC} \):    rC = 6.0 m.
Solution

a) The electric field is given by
\[ \bbox[#99CCFF,10px] {E=k_{0}\frac{q}{r^{2}}} \]
\[ \begin{gather} E_{B}=k_{0}\frac{Q}{r_{B}^{2}}\\ E_{B}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right)}{3.0^{2}}\\ E_{B}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right)}{9.0} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {E_{B}=-3\times 10^{3}\;\frac{\text{N}}{\text{C}}} \]
As the electric charge is negative, Q < 0, then it produces an electric field pointing inward toward the charge itself (Figure 1).
Figure 1

b) The electrostatic potential is given by
\[ \bbox[#99CCFF,10px] {V=k_{0}\frac{Q}{r}} \]
\[ \begin{gather} V_{B}=k_{0}\frac{Q}{r_{B}}\\ V_{B}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right)}{3.0} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {V_{B}=-9\times 10^{3}\;\text{V}} \]

c) If potential energy equal to zero at infinity, the potential energy at point C will be
\[ \bbox[#99CCFF,10px] {U=k_{0}\frac{Qq}{r}} \]
\[ \begin{gather} U_{C}=k_{0}\frac{Qq}{r_{C}}\\ U_{C}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right).\left(-5\times 10^{-9}\right)}{6.0}\\ U_{C}=\frac{135\times 10^{-6}}{6.0} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {U_{C}=2.3\times 10^{-5}\;\text{J}} \]

d) The work done by an external agent to bring a particle from infinity to point P,   \( {_{op}}{}{}{W}{_{\infty }^{P}} \),   is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {{_{op}}{}{}{W}{_{\infty }^{P}}=qV_{P}} \tag{I} \end{gather} \]
for the calculation of the work, we need to find the potential at point C,   VC
\[ \begin{gather} V_{C}=k_{0}\frac{Q}{r_{C}}\\ V_{C}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right)}{6.0}\\ V_{C}=-4.5\times 10^{3}\;\text{V} \end{gather} \]
substituting this value into the expression (I)
\[ \begin{gather} {_{op}}{}{}{W}{_{\infty}^{C}}=qV_{C}\\ {_{op}}{}{}{W}{_{\infty}^{C}}=\left(-5\times 10^{-9}\right)\times \left(-4.5\times 10^{3}\right) \end{gather} \]
\[ \bbox[#FFCCCC,10px] {{_{op}}{}{}{W}{_{\infty}^{C}}=2.3\times 10^{-5}\;\text{J}} \]

e) The electric charge Q produces at point C a field pointing inward, as well as calculated in item (a) to point B. The charge q has a negative value, q < 0, then the force that acts on this charge is towards the opposite direction to the field, it is a repulsive force (Figure 2).
As the charge moves in the opposite direction to the force, the work done by the electric field will be negative, and the same magnitude as the calculated in the previous item.

Figure 2
\[ {_{el}}{}{}{W}{_{\infty}^{C}}=-{_{op}}{}{}{W}{_{\infty }^{C}} \]
\[ \bbox[#FFCCCC,10px] {{_{el}}{}{}{W}{_{\infty}^{C}}=-2.3\times 10^{-5}\;\text{J}} \]

f) As the electric field is conservative, the work to move an electric charge between two points is independent of the path (in red in Figure 3) and depends only on the potential difference of the chosen points. The displacement in this case is against the direction of the force and the work done by the external agent will be positive \( {_{op}}{}{}{W}{_{C}^{A}} > 0 \).
\[ \begin{gather} {_{op}}{}{}{W}{_{C}^{A}}=q\left(V_{A}-V_{C}\right) \tag{II} \end{gather} \]
for the calculation of the work, we must find the value of the potential in A
\[ \begin{gather} V_{A}=k_{0}\frac{Q}{r_{A}}\\ V_{A}=9\times 10^{9}\times \frac{\left(-3\times 10^{-6}\right)}{1.0}\\ V_{A}=27\times 10^{3}\;\text{V} \end{gather} \]
substituting this value in equation (II) and the value of VC found earlier, we have the work to take a charge from point C to A, it will be
Figure 3
\[ \begin{gather} {_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[-27\times 10^{3}-\left(-4.5\times 10^{3}\right)\right]\\ {_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[\left(-27+4.5\right)\times 10^{3}\right]\\ {_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left(-22.5\times 10^{3}\right) \end{gather} \]
\[ \bbox[#FFCCCC,10px] {{_{op}}{}{}{W}{_{C}^{A}}=1,1.10^{-4}\;\text{J}} \]

g) As the charge moves in the opposite direction to the electric force the work of the electric field will be negative,   \( {_{el}}{}{}{W}{_{C}^{A}} < 0 \)
\[ \begin{gather} {_{el}}{}{}{W}{_{C}^{A}}=q\left(V_{C}-V_{A}\right)\\ {_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[-4.5\times 10^{3}-\left(-27\times 10^{3}\right)\right]\\ {_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[\left(-4.5+27\right)\times 10^{3}\right]\\ {_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left(22.5\times 10^{3}\right) \end{gather} \]
\[ \bbox[#FFCCCC,10px] {{_{el}}{}{}{W}{_{C}^{A}}=-1.1\times 10^{-4}\;\text{J}} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .