Solved Problem on Electric Potential

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

\[ \begin{gather} \bbox[#99CCFF,10px] {V=k_e\frac{Q_1}{d_1}+k_e\frac{Q_2}{d_2}+...+k_e\frac{Q_n}{d_n}} \end{gather} \]

a) For   \( Q_1=+6\times 10^{-6}\;\text C \) , \( Q_2=-6\times 10^{-6}\;\text C \)   and   \( d_1=d_2=d_{\small A}=0.3\;\text m \)

\[ \begin{gather} V=\left(9\times 10^9\;\frac{\mathrm{N.m^{\cancel 2}}}{\mathrm{C^{\cancel 2}}}\right)\left(\frac{6\times 10^{-6}\;\mathrm{\cancel C}}{0.3\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{{N.m^{\cancel 2}}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.3\;\mathrm{\cancel m}}\right)\\[5pt] V=\left(180\times 10^3\;\mathrm{\frac{N.m}{C}}\right)-\left(180\times 10^3\;\mathrm{\frac{N.m}{C}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V=0} \end{gather} \]

 

b) For   \( Q_1=Q_2=+6\times 10^{-6}\;\text C \)   and   \( d_1=d_2=d_{\small B}=0.1\;\text m \)

\[ \begin{gather} V=9\times 10^9\times\frac{6\times 10^{-6}}{0,1}+9\times 10^9\times\frac{6\times 10^{-6}}{0,1} \\[5pt] V=2\times 540\times 10^3 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V=1.08\times 10^6\;\text V} \end{gather} \]

 

c) For   \( Q_1=Q_2=-6\times 10^{-6}\;\text C \)   and   \( d_1=d_2=d_{\small C}=0.2\;\text m \)

\[ \begin{gather} V=\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.2\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{-6\times 10^{-6}\;\mathrm{\cancel C}}{0.2\;\mathrm{\cancel m}}\right)\\[5pt] V=\cancel 2\times\left(-\frac{54\times 10^{3}}{\cancel 2\times 10^{-1}}\;\frac{\mathrm{N.m}}{\mathrm C}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {V=-5.4\times 10^{5}\;\mathrm V} \end{gather} \]
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