JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Electrostatics

Charge

Q_{1}

and

Q_{2}

are at point

A

and

B

of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then

\frac{Q _{1}}{Q _{2}}

is proportional to:

Answer: (C)

Solution

Resolve the fields that

Q_{1}

and

Q_{2}

produce at

O

. The field due to

Q_{1}

at distance

x_{1}

has magnitude

\frac{k Q _{1}}{x _{1}^{2}}

and that due to

Q_{2}

at distance

x_{2}

\frac{k Q _{2}}{x _{2}^{2}}

, the two lying along the legs

O A

and

O B

of the right angle. For the resultant to be perpendicular to the hypotenuse

A B

, its component along

A B

must vanish, which gives

\frac{k Q _{2}}{x _{2}^{2}} cos α = \frac{k Q _{1}}{x _{1}^{2}} sin α

, where

cos α = \frac{x _{2}}{A B}

and

sin α = \frac{x _{1}}{A B}

. Substituting these,

\frac{Q _{2}}{x _{2}^{2}} \cdot x_{2} = \frac{Q _{1}}{x _{1}^{2}} \cdot x_{1}

, which simplifies to

\frac{Q _{2}}{x _{2}} = \frac{Q _{1}}{x _{1}}

. Hence

\frac{Q _{1}}{Q _{2}}

is proportional to

\frac{x _{1}}{x _{2}}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option