A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals:

Electrostatics PYQ — JEE Main 2009 · Shift 0 · Q62 with Solution

View written solutionFree

Correct answer: D

Set up the square

Let the side of the square be $a$ .

Place the charges at the corners as follows:

At $(0,0)$ : charge $Q$
At $(a,a)$ : charge $Q$ (opposite corner)
At $(a,0)$ : charge $q$
At $(0,a)$ : charge $q$

We need the net force on the charge $Q$ at $(0,0)$ to be zero.

Force on $Q$ due to the two charges $q$

Each charge $q$ is at distance $a$ from $Q$ .

So the magnitude of force due to each $q$ is

F_q = k\frac{|Qq|}{a^2}

These two forces act along the negative $x$ -axis and negative $y$ -axis respectively if $Q$ and $q$ are of same sign, and in opposite directions if signs differ. Since the net force must cancel the diagonal force due to the other $Q$ , $q$ must be of opposite sign to $Q$ .

Thus the force components due to the two $q$ charges are:

along $+x$ : magnitude $k\dfrac{|Qq|}{a^2}$
along $+y$ : magnitude $k\dfrac{|Qq|}{a^2}$

Hence their resultant is along the diagonal toward $(a,a)$ , with magnitude

F_{q,\text{net}} = \sqrt{\left(k\frac{|Qq|}{a^2}\right)^2 + \left(k\frac{|Qq|}{a^2}\right)^2} = \sqrt{2}\,k\frac{|Qq|}{a^2}

Force on $Q$ due to the opposite corner charge $Q$

Distance between opposite corners is

a\sqrt{2}

So the magnitude of force due to the other $Q$ is

F_Q = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2}

This force is along the diagonal away from $(a,a)$ if the two $Q$ charges have same sign. The resultant due to the two $q$ charges must oppose this, so this is consistent.

Condition for net force to be zero

For cancellation, the magnitudes along the same diagonal must be equal:

\sqrt{2}\,k\frac{|Qq|}{a^2} = k\frac{Q^2}{2a^2}

Cancel $k$ and $a^2$ :

\sqrt{2}|Qq| = \frac{Q^2}{2}

Assuming $Q\neq 0$ , divide by $|Q|$ :

\sqrt{2}|q| = \frac{|Q|}{2}

\frac{|Q|}{|q|} = 2\sqrt{2}

Since $q$ must be opposite in sign to $Q$ ,

\frac{Q}{q} = -2\sqrt{2}

Check options

The correct option is

\boxed{-2\sqrt{2}}

which is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

More from Electrostatics

View written solutionFree

Correct answer: D

Set up the square

Let the side of the square be $a$ .

Place the charges at the corners as follows:

At $(0,0)$ : charge $Q$
At $(a,a)$ : charge $Q$ (opposite corner)
At $(a,0)$ : charge $q$
At $(0,a)$ : charge $q$

We need the net force on the charge $Q$ at $(0,0)$ to be zero.

Force on $Q$ due to the two charges $q$

Each charge $q$ is at distance $a$ from $Q$ .

So the magnitude of force due to each $q$ is

F_q = k\frac{|Qq|}{a^2}

Thus the force components due to the two $q$ charges are:

along $+x$ : magnitude $k\dfrac{|Qq|}{a^2}$
along $+y$ : magnitude $k\dfrac{|Qq|}{a^2}$

Hence their resultant is along the diagonal toward $(a,a)$ , with magnitude

F_{q,\text{net}} = \sqrt{\left(k\frac{|Qq|}{a^2}\right)^2 + \left(k\frac{|Qq|}{a^2}\right)^2} = \sqrt{2}\,k\frac{|Qq|}{a^2}

Force on $Q$ due to the opposite corner charge $Q$

Distance between opposite corners is

a\sqrt{2}

So the magnitude of force due to the other $Q$ is

F_Q = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2}

This force is along the diagonal away from $(a,a)$ if the two $Q$ charges have same sign. The resultant due to the two $q$ charges must oppose this, so this is consistent.

Condition for net force to be zero

For cancellation, the magnitudes along the same diagonal must be equal:

\sqrt{2}\,k\frac{|Qq|}{a^2} = k\frac{Q^2}{2a^2}

Cancel $k$ and $a^2$ :

\sqrt{2}|Qq| = \frac{Q^2}{2}

Assuming $Q\neq 0$ , divide by $|Q|$ :

\sqrt{2}|q| = \frac{|Q|}{2}

\frac{|Q|}{|q|} = 2\sqrt{2}

Since $q$ must be opposite in sign to $Q$ ,

\frac{Q}{q} = -2\sqrt{2}

Check options

The correct option is

\boxed{-2\sqrt{2}}

which is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

Electrostatics question

Electrostatics question

More from Electrostatics

Electrostatics question

Electrostatics question

More from Electrostatics