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).
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.
\[
{_{el}}{}{}{W}{_{\infty}^{C}}=-{_{op}}{}{}{W}{_{\infty }^{C}}
\]
\[ \bbox[#FFCCCC,10px]
{{_{el}}{}{}{W}{_{\infty}^{C}}=-2.3\times 10^{-5}\;\text{J}}
\]
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
\[
\begin{gather}
{_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[-27\times 10^{3}-\left(-4.5\times 10^{3}\right)\right]\\
{_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[\left(-27+4.5\right)\times 10^{3}\right]\\
{_{op}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left(-22.5\times 10^{3}\right)
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{{_{op}}{}{}{W}{_{C}^{A}}=1,1.10^{-4}\;\text{J}}
\]
g) As the charge moves in the opposite direction to the electric force the work of the electric field will
be negative,
\( {_{el}}{}{}{W}{_{C}^{A}} < 0 \)
\[
\begin{gather}
{_{el}}{}{}{W}{_{C}^{A}}=q\left(V_{C}-V_{A}\right)\\
{_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[-4.5\times 10^{3}-\left(-27\times 10^{3}\right)\right]\\
{_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left[\left(-4.5+27\right)\times 10^{3}\right]\\
{_{el}}{}{}{W}{_{C}^{A}}=\left(-5\times 10^{-9}\right)\times \left(22.5\times 10^{3}\right)
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{{_{el}}{}{}{W}{_{C}^{A}}=-1.1\times 10^{-4}\;\text{J}}
\]