Question

Four charges +q, +2q, +q and -2q are placed at the corner of square ABCD respectively. The force on unit positive charge kept at the centre 'O' is

• A. along the diagonal AC
• B. Zero
• C. perpendicular to AD
• D. along the diagonal BD

Answer 1

The Correct Option is D

Consider the following charges placed at the corners of a square:

• +q at corner A
• +2q at corner B
• +q at corner C
• -2q at corner D

We need to find the net force on the unit positive charge placed at the center of the square (O).

Step 1: Coulomb's Law

The force between two charges is given by Coulomb's law:

\( F = \frac{k |q_1 q_2|}{r^2} \)

Step 2: Analyzing the forces exerted by each charge on the unit positive charge at O

• Force exerted by +q at A: The force will be attractive and directed along the line AO, from A to O.
• Force exerted by +2q at B: The force will be attractive and directed along the line BO, from B to O.
• Force exerted by +q at C: The force will be attractive and directed along the line CO, from C to O.
• Force exerted by -2q at D: The force will be repulsive and directed along the line DO, from D to O.

Step 3: Direction of the forces:

Since the charges at A, B, and C are positive, they attract the unit positive charge at O. The charge at D is negative and repels the unit positive charge. Therefore, the forces along the lines AO, BO, and CO have equal magnitudes but are mutually perpendicular, directed towards the center of the square.

The force along the line DO is repulsive and directed away from the charge at D, and it has a different magnitude from the forces along AO, BO, and CO.

Step 4: Vector Addition of Forces

The forces along AO, BO, and CO are mutually perpendicular. Since they have equal magnitudes and are directed at 90° to each other, they will cancel out each other when added vectorially.

The force along the line DO will not cancel out completely. It will have a net component along the diagonal BD of the square. The net force on the unit positive charge at O is along the diagonal BD.

Therefore, the net force on the unit positive charge at the center O is along the diagonal BD (option D).

Answer 2

Let the square be ABCD with side length \( a \). The charges are placed at the corners as follows:

• A: \(+q\)
• B: \(+2q\) 
• C: \(+q\)
• D: \(-2q\)

A unit positive charge (\(+1\)) is placed at the center 'O'.

The distance of the center 'O' from each corner (A, B, C, D) is the same. Let this distance be \( r \).

The length of the diagonal AC = BD = \( \sqrt{a^2 + a^2} = a\sqrt{2} \).

The distance \( r \) is half the diagonal length: \( r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} \).

Therefore, \( r^2 = \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{a^2}{2} \).

Now, let's calculate the force exerted by each corner charge on the unit positive charge at O using Coulomb's Law, \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k \) is Coulomb's constant.

Force due to charge at A (+q):

\( F_A = k \frac{|(+q)(+1)|}{r^2} = k \frac{q}{a^2/2} = \frac{2kq}{a^2} \)

The force is repulsive, directed along AO (away from A), which is along the line OC.

Force due to charge at B (+2q):

\( F_B = k \frac{|(+2q)(+1)|}{r^2} = k \frac{2q}{a^2/2} = \frac{4kq}{a^2} \)

The force is repulsive, directed along BO (away from B), which is along the line OD.

Force due to charge at C (+q):

\( F_C = k \frac{|(+q)(+1)|}{r^2} = k \frac{q}{a^2/2} = \frac{2kq}{a^2} \)

The force is repulsive, directed along CO (away from C), which is along the line OA.

Force due to charge at D (-2q):

\( F_D = k \frac{|(-2q)(+1)|}{r^2} = k \frac{2q}{a^2/2} = \frac{4kq}{a^2} \)

The force is attractive, directed along OD (towards D), which is along the line OD.

Now, let's find the net force by vector addition.

Consider the forces along the diagonal AC:

• \( \vec{F}_A \) is along OC.
• \( \vec{F}_C \) is along OA.

Since \( |\vec{F}_A| = |\vec{F}_C| = \frac{2kq}{a^2} \) and they are in opposite directions (\( \vec{OA} = -\vec{OC} \)), their vector sum is zero:

\( \vec{F}_{AC} = \vec{F}_A + \vec{F}_C = 0 \)

Consider the forces along the diagonal BD:

• \( \vec{F}_B \) is along OD.
• \( \vec{F}_D \) is along OD.

Both forces \( \vec{F}_B \) and \( \vec{F}_D \) are in the same direction (along OD). Their magnitudes are \( |\vec{F}_B| = \frac{4kq}{a^2} \) and \( |\vec{F}_D| = \frac{4kq}{a^2} \).

The resultant force along the diagonal BD is:

\( \vec{F}_{BD} = \vec{F}_B + \vec{F}_D \)

The magnitude is \( |\vec{F}_{BD}| = |\vec{F}_B| + |\vec{F}_D| = \frac{4kq}{a^2} + \frac{4kq}{a^2} = \frac{8kq}{a^2} \).

The direction is along OD.

The total net force \( \vec{F}_{net} \) on the charge at O is the vector sum of the forces along the diagonals:

\( \vec{F}_{net} = \vec{F}_{AC} + \vec{F}_{BD} = 0 + \vec{F}_{BD} = \vec{F}_{BD} \)

So, the net force is \( \frac{8kq}{a^2} \) directed along OD.

The direction OD lies along the diagonal BD.