Solved Problem on Electric Potential

A point charge of 200 μC moves from point A to point B in an electric field, the work done by the electric force is 8×10−4 J. Calculate:
a) The potential difference between points A and B;
b) The electric potential taking B as a reference.

 

Problem data:

  • Electric charge:    q = 200 μC = 200×10−6 C;
  • Work done by electric force:    \( {_{{\small F}_{\small E}}}W{_a^b}=8\times 10^{-4}\;\text J \) .

Problem diagram:

Figure 1

Solution:

a) The work of electric force as a function of the charge and the potential difference is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{_{{\small F}_{\small E}}}W{_a^b}=q(V_a-V_b)} \end{gather} \]
\[ \begin{gather} V_a-V_b=\frac{{_{{\small F}_{\small E}}}W{_a^b}}{q} \\[5pt] V_a-V_b=\frac{8\times 10^{-4}\times 10^{-2}\times 10^6\;\mathrm{\cancel C .V}}{2\;\mathrm{\cancel C}} \\[5pt] V_a-V_b=\frac{\cancelto{4}{8}\times 10^{-4}}{\cancel 2\times 10^2\times 10^{-6}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V_a-V_b=4\;\text V} \end{gather} \]

 

b) We chose the potential of point B as a reference. The potential of point B shall be zero, Vb = 0. Substituting this value into the expression found in the previous item, the potential will be

\[ \begin{gather} V_a=4-0 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V_a=4\;\text V} \end{gather} \]
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