Question

Four point charges qa=2C, qb=-5 C, qc=2C, qd=-5 C, are placed along the corners of a square ABCD having sides 10 cm each. Find the force acting on 1 C charge placed at the centre of the square​

Answer 1

Answer:

ANSWER

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

Where,

(Sides) AB=BC=CD=AD=10cm

(Diagonals) AC=BD=10

2

cm

AO=OC=DO=OB=5

2

cm

A charge of amount 1μC is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1μC charge at centre O is zero.

Answer 2

Answer:

Net force on 1μC charge = 0

Explanation:

A rough diagram is shown in figure.

here ABCD is a square in which charges are placed in each corners as shown in figure.

Now, force due to B on 1μC FOB = Force due to D on 1μC, FOD

And both act opposite directions so, these forces are cancelled.

Similarly, force due to A on 1μC, FAO= force due to C on 1μC, FCO

and both act opposite directions so, thses forces are also cancelled.

Hence, net force acts on 1μC = 0