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(D) Given: $q_1 = +16 \mu C$,$q_2 = -9 \mu C$,distance $d = 10 \ cm$.Let the point where the resultant electric field is zero be at a distance $x$ fro

Vedclass · Accepted Answer

$30$

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(D) Given: $q_1 = +16 \mu C$,$q_2 = -9 \mu C$,distance $d = 10 \ cm$.Let the point where the resultant electric field is zero be at a distance $x$ from the $-9 \mu C$ charge. Since the charges are unlike,the null point lies outside the region between the charges,closer to the charge with the smaller magnitude.At the null point,the magnitude of the electric field due to $q_1$ must equal the magnitude of the electric field due to $q_2$.$E_1 = E_2$$\frac{k |q_1|}{(d + x)^2} = \frac{k |q_2|}{x^2}$$\frac{16}{(10 + x)^2} = \frac{9}{x^2}$Taking the square root on both sides:$\frac{4}{10 + x} = \frac{3}{x}$$4x = 3(10 + x)$$4x = 30 + 3x$$x = 30 \ cm$Thus,the resultant electric field is zero at a distance of $30 \ cm$ from the $-9 \mu C$ charge.

Two point unlike charges having magnitudes $+16 \mu C$ and $-9 \mu C$ are separated by a distance of $10 \ cm$ in air. The resultant electric field will be zero at a distance of . . . . . . from the $-9 \mu C$ charge. (in $cm$)

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