Solved Problem on Electric Potential
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Determine the electric potential at point P in cases (A), (B), and (C) of the figure below, Q = 6 μC, \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \) , and given the distances of the charges to point P:
a) dA = 0.3 m;
b) dB = 0.1 m;
c) dC = 0.2 m.
Problem data:
- Charge: Q=6 μC=6×10−6 C;
- Coulomb Constant: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \) .
Solution
The electric potential at a point, due to various charges, is given by the algebraic sum of the potential of each charge
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V=k_e\frac{Q_1}{d_1}+k_e\frac{Q_2}{d_2}+...+k_e\frac{Q_n}{d_n}}
\end{gather}
\]
a) For \( Q_1=+6\times 10^{-6}\;\mathrm C \) , \( Q_2=-6\times 10^{-6}\;\mathrm C \) and \( d_1=d_2=d_{\small A}=0.3\;\mathrm m \)
\[
\begin{gather}
V=\left(9\times 10^9\;\frac{\mathrm{N.m^{\cancel 2}}}{\mathrm{C^{\cancel 2}}}\right)\left(\frac{6\times 10^{-6}\;\mathrm{\cancel C}}{0.3\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{{N.m^{\cancel 2}}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.3\;\mathrm{\cancel m}}\right)\\[5pt]
V=\left(180\times 10^3\;\mathrm{\frac{N.m}{C}}\right)-\left(180\times 10^3\;\mathrm{\frac{N.m}{C}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{V=0}
\end{gather}
\]
b) For \( Q_1=Q_2=+6\times 10^{-6}\;\mathrm C \) and \( d_1=d_2=d_{\small B}=0.1\;\mathrm m \)
\[
\begin{gather}
V=\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{6\times 10^{-6}\;\mathrm{\cancel C}}{0.1\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{6\times 10^{-6}\;\mathrm{\cancel C}}{0.1\;\mathrm{\cancel m}}\right)\\[5pt]
V=2\times\left(540\times 10^3\;\mathrm{\frac{N.m}{C}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{V=1.08\times 10^6\;\mathrm V}
\end{gather}
\]
c) For \( Q_1=Q_2=-6\times 10^{-6}\;\mathrm C \) and \( d_1=d_2=d_{\small C}=0.2\;\mathrm m \)
\[
\begin{gather}
V=\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.2\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.2\;\mathrm{\cancel m}}\right)\\[5pt]
V=\cancel 2\times\left(-\frac{54\times 10^{3}}{\cancel 2\times 10^{-1}}\;\frac{\mathrm{N.m}}{\mathrm C}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{V=-5.4\times 10^{5}\;\mathrm V}
\end{gather}
\]
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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 .