Melia40idn
Answered

IDNLearn.com: Your trusted source for finding accurate answers. Ask your questions and receive reliable and comprehensive answers from our dedicated community of professionals.

Two identical metallic sphere having unequal opposite charges are are placed
distance of 0.05m apart in air.
After bringing them in contact
with each other, they are again placed at the same distance apart, now the force of repulsion between them is 0.108 N. Calculate the final charge on each of them.​

Sagot :

Gauravkumar952816idn

Answer:

Let the initial charges be q1 and q2 respectively.

After they come in contact, the charges are rearranged such that they acquire same charge.

let us say that charge on each of them is Q.

They are again brought apart at a distance of 0.9 m. Hence, the force between them will be given as

F = kQ2 / r2

0.025 = (9×109 x Q2) / 0.92

Q2 = 0.025 x 0.92 / 9×109

Q = 1.5 x 10-6 C

Explanation:

hey buddy, ❤can u plz subscribe to my UTube channel gtron9528 plz

We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.