Question

Two charges of 1.0 C each are placed
one meter apart in free space. The force
between them will be.
(A) 1.0N (B) 9x10°N
(C) 9x10-'N (D) 10N​

Answer 1

Answer:

$9 \times {10}^{9} \: newtons$

Explanation:

F=kq1q2/r^2

=9×10^9(1)(1)/1

=9×10^9N

Answer 2

Answer:

$F=9\times 10^9\ N$

Explanation:

Given that

q₁=q₂= 1 C

Distance ,r= 1 m

We know that force between two charge given as

$F=K\dfrac{q_1q_2}{r^2}$

Where K= Constant

Now by putting the values

$F=K\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{1\times 1}{1^2}\ N$

$F=9\times 10^9\ N$

We know that two same charge have repulsive force and two opposite charge have attractive force.

Here Force is repulsive in nature.

$F=9\times 10^9\ N$