Question

two spherical carrying charges 6 microcoulomb and 9 microcoulomb separated by distance di and explains s e and forth and repulsion unit charge of - 3 microcoulomb are given to the both of the spear and kept at the same distance as before the new force of repulsion is

Answer 1

before
F=k(6)(9)/d raise to 2
then
F'=k(3)(6)/d raise to 2
F'/F=1/3
SO


F'=F/3

Answer 2

Given:

Two charges 6μC and 9 μC separated by distance d.

Additional charge -3μC is added which makes the total charge on first as 3μC and on second 6μC. The distance remains the same.

To find: The new force F' between the two charges.

Formula Used:

According to Coulomb's force:

$F =k\frac{qQ}{d^2}$

Where, k is the Coulomb's constant, q and Q are two charges having d distance between them.

Calculations:

Take the ratio:

$\frac{F'}{F} = \frac{3\times 6}{6\times 9} \\F'= \frac{1}{3}F$

The new force is one-third the original force.

Since, the distance is not given, it cannot be calculated exactly.

Learn more about: Coulomb's force

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