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(C) The magnitude of the electric field is proportional to the density of the electric field lines.Let $E_A$ and $E_B$ be the electric field magnitude

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) The magnitude of the electric field is proportional to the density of the electric field lines.Let $E_A$ and $E_B$ be the electric field magnitudes at points $A$ and $B$ respectively.Let $d_A$ and $d_B$ be the spacing between the field lines at $A$ and $B$ respectively.From the figure,the spacing at $A$ is $x$ and the spacing at $B$ is $y$.We are given $y \approx 2x$.The density of field lines is inversely proportional to the spacing between them.Therefore,$\frac{E_A}{E_B} = \frac{d_B}{d_A} = \frac{y}{x}$.Substituting the given values,$\frac{40}{E_B} = \frac{2x}{x} = 2$.Thus,$E_B = \frac{40}{2} = 20 \ N/C$.

The figure shows the electric field lines. The spacing between the lines is parallel to the paper at every point. If the magnitude of the field at $A$ is $40 \ N/C$,then the approximate magnitude of the field at $B$ is ....... $N/C$.

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