CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Four charges each equal to Q are placed at the four corners of a square and a charge q is placed at the centre of the square. If the system is in equilibrium, the value of q is :

A
Q2(1+22)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Q4(1+22)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Q4(1+22)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Q2(1+22)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B Q4(1+22)


Step 1 : Calculate all the required distances between the charges :
Let the side length of the square be a.

Refer Figure, In ACD,
AD2+CD2=AC2

a2+a2=AC2 AC=2a

However, OC=AC2 =a2

Step 2 : Equilibrium of central charge
By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.
Hence, Net force on central charge will always be zero, irrespective of value of charge q.

Therefore, to find value of q we have to check equilibrium of any one charge at the corner.

Step 3 : Force due to all the charges at point C
Force on charge at C due to B, F1=KQ2a2

Force on charge at C due to D, F2=KQ2a2

Force on charge at C due to A, F4=KQ2AC2=KQ22a2

Force on charge at C due to q at centre, F3=KqQOC2=2KqQa2

Step 4 : Apply the equilibrium condition at C :

For the system to be in equilibrium, net force acting on charge at C must be zero.

So, F3+F1+F2+F4=0
Resultant of F1 and F2 (Along OC, by symmetry) = F21+F22=2F1 Since (|F1|=|F2|)

Also, F3 & F4 are along OC, Therefore, magnitudes of sum of these forces should be zero.

|F3|+2|F1|+|F4|=0

2KqQa2 +2KQ2a2 +KQ22a2=0

2q=(2Q+Q2)

q=Q4(1+22)

2102234_467912_ans_94125d4a53104c73b2803857c2c974db.png

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intensity vs Photocurrent
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon