Question

Two equal positive charges kept at distance – repel each other with a certain force. The force of repulsion
remains the same even when the magnitude of one charge is reduced by 1 uc and that of the other is increased
2 uc. Find the two charges.​

Answer 1

Answer:

Explanation:

F1 = k * q1 * Q2 / d ^ 2

ایف 1 کو فاصلے کیلئے طاقت بننے دیں

اب اگر آدھا رہ جاتا ہے تو ، مساوات ہوگی ،

ایف 1 = کے * کیو 1 * کیو 2 / (ڈی / 2) ^ 2

= 4 * کے * کیو 1 * کیو 2 / ڈی ^ 2

کونسا،

F = 4F1

جب فاصلہ آدھا ہوجائے تو F وہ قوت ہے

F = 4F

The equation for force is

F1=k*q1*q2/d^2

Let F1 be the force for distance d

Now if d is halved,the equation will be,

F1=k*q 1*q2/(d/2)^2

=4*k*q1*q2/d^2

Which is,

F=4F1

Where F is the force when distance is halved.

F=4F