Question

Two charge of opposite nature are placed at some distance. If the force of attraction between them is 0.108N.After connecting two charge with a wire the force of repulsion becomes 0.036N.If the seperation between charge in both case is 50cm.Find the initial charge value? ​

Answer 1

Answer:

Let initial charge be Q1 and Q2.

Initial force =F=

r

2

K×Q1×Q2

=−0.108

F=

0.25

9×10

9

×Q1×Q2

=−0.108

Q1.Q2=−3×10

−12

Now on connecting them, since they are identical so they will have equal charges.

Final charge =

2

Q1+Q2

Final force =F=

4×0.25

9×10

9

×(Q1+Q2)

2

=0.036

Q1+Q2=±2×10

−6

Q2=±1.0×10

−6

C,Q1=∓3.0×10

−6

C