Solved Problem on Electric Potential

An electric field is produced in a vacuum by two point charges of −2 μC and 5 μC. Calculate:
a) The electric potential, at a point P, which is 0.2 m from the first charge and 0.5 m from the second;
b) The electric potential energy that a charge q = 6×10⁻⁸ C acquires when placed at P.

Problem data:

Electric charge 1: q₁ = −2 μC = −2×10⁻⁸ C;
Distance from P to charge 1: d₁ = 0.2 m;
Electric charge 2: q₂ = 5 μC = 5×10⁻⁸ C;
Distance from P to charge 2: d₂ = 0.5 m;
Coulomb Constant in a vacuum: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \) .

Solution

a) The electric potential at a point due to various charges, is given by the algebraic sum of the potential of each charge

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

for the two charges q₁ and q₂

\[ \begin{gather} V=\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{-2\times 10^{-6}\;\mathrm{\cancel C}}{0.2\;\mathrm{\cancel m}}\right)+\left(9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^{\cancel 2}}}\right)\left(\frac{5\times 10^{-6}\;\mathrm{\cancel C}}{0.5\;\mathrm{\cancel m}}\right)\\[5pt] V=-\left(\frac{9\times 10^9\times\cancel 2\times 10^{-6}}{\cancel 2\times 10^{-1}}\;\mathrm{\frac{N.m}{C}}\right)+\left(\frac{9\times 10^9\times\cancel 5\times 10^{-6}}{\cancel 5\times 10^{-1}}\;\mathrm{\frac{N.m}{C}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V=0} \end{gather} \]

b) The electric potential energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {U=qV} \end{gather} \]

\[ \begin{gather} U=\left(6\times 10^{-8}\;\mathrm C\right)\times 0 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {U=0} \end{gather} \]

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