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

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, V_B = 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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