Solved Problem on Electric Potential
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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:    \( {_{{\small F}_{\small E}}}W{_{\small A}^{\small B}}=8\times 10^{-4}\;\mathrm{J} \) .
Problem diagram:

Figure 1

Solution

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

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


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
\[ \begin{gather} V_{\small A}=4\;\mathrm V-0 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V_{\small A}=4\;\mathrm V} \end{gather} \]
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