Question

The force of attraction between two oppositely charged pith balls is 5.3 × 10-6 newtons. If the charge on the two objects is 6.7 × 10-9 coulombs, what is the distance between the two charges? (k = 9.0 × 109 newton·meter2/coulombs2)
A.
2.76 × 10-1 meters

B.
7. 0 × 10-1 meters

C.
6.7 × 10-1 meters

D.
8.2 × 10-1 meters

E.
9.6 × 10-1 meters

Answer 1

your answer is E. 9.6*10-1

hope this helped you

Answer 2

Answer:

A.  2.76 × 10-1 meters

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the distance between the charges

In this problem, we know:

$F=5.3 \cdot 10^{-6} N$

$q_1 = 1_2 = 6.7 \cdot 10^{-9} C$

$k=9.0 \cdot 10^9 N/m^2 C^2$

So we can re-arrange the equation to find the distance between the two charges:

$r=\sqrt{\frac{k q_1 q_2}{F}}=\sqrt{\frac{9.0\cdot 10^9 (6.7 \cdot 10^{-9})^2}{5.3\cdot 10^{-6}}}=0.276 m$

Which corresponds to option A, 2.76 × 10-1 meters.