Chapter 14, Problem 1RQ

To determine

Choose the correct option that provides the required point on an equipotential line.

Answer to Problem 1RQ

The correct option is $\underset{\_}{\left(\text{a}\right)\left(1.04,2.03,0\right)}$.

Explanation of Solution

Calculation:

Consider the given expression for the electric field.

$E=0.3a_x-0.4a_y\text{V/m}$        (1)

Write the general expression for the electric field.

$E=E_x{a}_x+E_y{a}_y\text{V/m}$        (2)

By comparing Equation (1) and (2), the values of $E_x\text{and}{E}_y$ are 0.3 and $-0.4$ respectively.

Consider that the movements $\Delta x\text{and}\Delta y$ in the x- and y-directions.

Calculate $\Delta x$.

$\Delta x=\frac{-\Delta I{E}_y}{\sqrt{E_x^2+E_y^2}}$

Substitute 0.05 m for $\Delta I$, 0.3 for $E_x$, and $-0.4$ for $E_y$.

$\begin{array}{c}\Delta x=\frac{-\left(0.05\text{m}\right)\left(-0.4\right)}{\sqrt{{\left(0.3\right)}^2+{\left(-0.4\right)}^2}}\\ =0.04\end{array}$

Calculate $\Delta y$.

$\Delta y=\frac{\Delta I{E}_x}{\sqrt{E_x^2+E_y^2}}$

Substitute 0.05 m for $\Delta I$, 0.3 for $E_x$, and $-0.4$ for $E_y$.

$\begin{array}{c}\Delta y=\frac{\left(0.05\text{m}\right)\left(0.3\right)}{\sqrt{{\left(0.3\right)}^2+{\left(-0.4\right)}^2}}\\ =0.03\end{array}$

Calculate the point $\left(x+\Delta x,y+\Delta y,0\right)$.

$\text{Point}=\left(x+\Delta x,y+\Delta y,0\right)$

Substitute 1 for x, 2 for y, 0.04 for $\Delta x$, and 0.03 for $\Delta y$.

$\begin{array}{c}\text{Point}=\left(1+0.04,2+0.03,0\right)\\ =\left(1.04,2.03,0\right)\end{array}$

Conclusion:

Thus, the correct option is $\underset{\_}{\left(\text{a}\right)\left(1.04,2.03,0\right)}$.