Question

calculate electric field strength at (2,3)centimetre due to a charge of 100 mc from origin?​

Answer 1

Answer:

Solution: Suppose that the line from $q_a$ to $q_b$ runs along the $x$-axis. It is clear, from Coulomb's law, that the electrostatic force exerted on any charge placed on this line is parallel to the $x$-axis. Thus, the electric field at any point along this line must also be aligned along the $x$-axis. Let the $x$-coordinates of charges $q_a$ and $q_b$ be $-c/2$ and $+c/2$, respectively. It follows that the origin ($x=0$) lies halfway between the two charges. The electric field $E_a$ generated by charge $q_a$ at the origin is given by

\begin{displaymath}

E_a = k_e\,\frac{q_a}{(c/2)^2} = (8.988\times 10^9)\, \frac{(50\times 10^{-6})}{(0.5)^2}

= 1.80\times 10^6\,{\rm N\,C}^{-1}.

\end{displaymath}

The field is positive because it is directed along the $+x$-axis (i.e., from charge $q_a$ towards the origin). The electric field $E_b$ generated by charge $q_b$ at the origin is given by

\begin{displaymath}

E_b = -k_e\,\frac{q_b}{(c/2)^2} = -(8.988\times 10^9) \,\fra...

...\times 10^{-6})}{(0.5)^2}

= -3.60\times 10^6\,{\rm N\,C}^{-1}.

\end{displaymath}

The field is negative because it is directed along the $-x$-axis (i.e., from charge $q_b$ towards the origin). The resultant field $E$ at the origin is the algebraic sum of $E_a$ and $E_b$ (since all fields are directed along the $x$-axis). Thus,

\begin{displaymath}

E = E_a + E_b = -1.8 \times 10^6\,{\rm N\,C}^{-1}.

\end{displaymath}

Since $E$ is negative, the resultant field is directed along the $-x$-axis.

The force $f$ acting on a charge $q_c$ placed at the origin is simply

\begin{displaymath}

f = q_c\,E = (20\times 10^{-6})\,(-1.8\times 10^6)=-36 \,{\rm N}.

\end{displaymath}

Since $f<0$, the force is directed along the $-x$-axis.

Explanation:

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