Two point charges -5 mu C and +5 mu C are pla | vedclass

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Question

(A) Let the charges be $q_A = -5 \mu C$ and $q_B = +5 \mu C$. Let the distance $AB = d = 5 \ cm$. Let point $C$ be at a distance $x$ from $A$ and $y$

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$AC = BC$

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(A) Let the charges be $q_A = -5 \mu C$ and $q_B = +5 \mu C$. Let the distance $AB = d = 5 \ cm$. Let point $C$ be at a distance $x$ from $A$ and $y$ from $B$. For the resultant electric field at $C$ to be parallel to the line $AB$,the vertical components of the electric fields produced by $q_A$ and $q_B$ must cancel each other out.Let $	heta_A$ and $	heta_B$ be the angles made by the vectors $\vec{E}_A$ and $\vec{E}_B$ with the line $AB$. For the vertical components to cancel,we must have $E_A \sin 	heta_A = E_B \sin 	heta_B$.Since the magnitudes of the charges are equal $(|q_A| = |q_B| = q)$,the field magnitudes are $E_A = \frac{kq}{x^2}$ and $E_B = \frac{kq}{y^2}$.Thus,$\frac{kq}{x^2} \sin 	heta_A = \frac{kq}{y^2} \sin 	heta_B$. From the geometry of the triangle $ABC$,$\sin 	heta_A = \frac{h}{x}$ and $\sin 	heta_B = \frac{h}{y}$,where $h$ is the perpendicular distance of $C$ from line $AB$.Substituting these,we get $\frac{kq}{x^2} \cdot \frac{h}{x} = \frac{kq}{y^2} \cdot \frac{h}{y}$,which simplifies to $\frac{1}{x^3} = \frac{1}{y^3}$,implying $x = y$.Therefore,$AC = BC$.

Two point charges $-5 \mu C$ and $+5 \mu C$ are placed at $A$ and $B$ respectively on a straight line separated by $5 \ cm$. At a point $C$,if the resultant electric field is parallel to the line joining the points $A$ and $B$,then:

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