Solved Problem on Electric Field
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The graph represents the magnitude of the electric field produced by a positive point charge Q
as a function of the distance to the charge. The Coulomb Constant in a vacuum is
\( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
Determine:
a) The charge Q;
b) The magnitude of the electric force that acts on a charge q = −1×10−5 C, placed at 2 m of Q;
c) The magnitude of the electric force that acts on a charge q = 1×10−5 C, placed at 1 m of Q.
a) The charge Q;
b) The magnitude of the electric force that acts on a charge q = −1×10−5 C, placed at 2 m of Q;
c) The magnitude of the electric force that acts on a charge q = 1×10−5 C, placed at 1 m of Q.
Problem data:
- Electric charge in situation 1: q1 = −1×10−5 C;
- Distance from charge Q up to charge q1: d1 = 2 m;
- Electric charge in situation 2: q2 = 1×10−5 C;
- Distance from charge Q up to charge q2: d2 = 1 m;
- Coulomb Constant in a vacuum: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
a) The magnitude of the electric field is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{E=k_e\frac{q}{r^2}}
\end{gather}
\]
From the graph, at a distance of r = 1 m from the charge Q, the electric
field is equal to E = 18×103 N/C, substituting this data, we
find the charge Q
\[
\begin{gather}
E=k_e\frac{Q}{r^2}\\[5pt]
Q=\frac{Er^2}{k_e}\\[5pt]
Q=\frac{\left(18\times 10^3\;\mathrm{\frac{\cancel N}{\cancel C}}\right)\left(1\;\mathrm{\cancel{m^2}}\right)}{9\times 10^9\;\mathrm{\frac{\cancel N.\cancel{m^2}}{C^{\cancel 2}}}}\\[5pt]
Q=2\times 10^3\times 10^{-9}\;\mathrm C
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{Q=2\times 10^{-6}\;\mathrm C}
\end{gather}
\]
b) Using Coulomb's Law, the magnitude of the electric force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{\small E}=k_e\frac{|\;q_1\;||\;q_2\;|}{r^2}} \tag{I}
\end{gather}
\]
substituting the charge found in item (a) and r = d1
\[
\begin{gather}
F_{\small E}=k_e\frac{|\;Q\;||\;q_{1}\;|}{d_{1}^{2}}\\[5pt]
F_{\small E}=\left(9\times 10^{9}\;\mathrm{\small{\frac{N.m^2}{C^2}}}\right)\frac{\left(2\times 10^{-6}\;\mathrm C\right)\left(1\times 10^{-5}\;\mathrm C\right)}{\left(2\;\mathrm m\right)^2}\\[5pt]
F_{\small E}=\left(9\times 10^{9}\;\mathrm{\small\frac{N.\cancel{m^2}}{\cancel{C^2}}}\right)\left(\frac{\cancel 2\times 10^{-6}\times 10^{-5}\;\mathrm{\cancel{C^2}}} {2^{\cancel 2}\;\mathrm{\cancel{m^2}}}\right)\\[5pt]
F_{\small E}=4.5\times10^9\times 10^{-11}\;\mathrm N
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{F_{\small E}=4.5\times 10^{-2}\;\mathrm N}
\end{gather}
\]
c) Applying the equation (I) and substituting the charge found in item (a) and r = d2
\[
\begin{gather}
F_{\small E}=k_e\frac{|\;Q\;||\;q_{2}\;|}{d_{2}^{2}}\\[5pt]
F_{\small E}=\left(9\times 10^9\;\mathrm{\small\frac{N.m^2}{C^2}}\right)\frac{\left(2\times 10^{-6}\;\mathrm C\right)\left(1\times 10^{-5}\;\mathrm C\right)}{\left(1\;\mathrm m\right)^2}\\[5pt]
F_{\small E}=\left(9\times 10^9\;\mathrm{\small\frac{N.\cancel{m^2}}{\cancel{C^2}}}\right)\left(\frac{2\times 10^{-6}\times 10^{-5}\;\mathrm{\cancel{C^2}}} {1\;\mathrm{\cancel{m^2}}}\right)\\[5pt]
F_{\small E}=18\times 10^{9}\times 10^{-11}\;\mathrm N
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{F_{\small E}=18\times 10^{-2}\;\mathrm N}
\end{gather}
\]
Note: the charge 1 is negative, the force acts in the opposite direction of the electric
field, and in the direction of the charge Q, opposite charges attract each other. Charge 2 is
positive, and the force acts in the same direction of the electric field, in the direction away from the
charge Q, like charges repel each other.
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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 .