Question

The force of repulsion between two like–charged table tennis balls is 8.2 × 10-7 newtons. If the charge on the two objects is 6.7 × 10-9 coulombs each, what is the distance between the two charges?
(k = 9.0 × 109 newton·meters2/coulomb2)

A.
0.32 meters
B.
0.70 meters
C.
6.7 meters
D.
8.2 meters

Answer 1

Answer:

the answer is b for plato users

Explanation:

Answer 2

solution

In This question we have given ,

Charge on each ball=$q=6.7\times 10^{-9} C$

force of repulsion between balls is $F=8.2\times 10^{-7}N$

Let distance between ball be x

We know by Coulombs law,

$F=\frac{k\times q_{1}q_{2}}{x^2}$..............(1)

here,$k = 9.0\times 10^9 Nm^2C^{-2}$

Put values of k and charges in equation 1

$8.2\times 10^{-7}N=\frac{9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}{x^2}$

$8.2\times 10^{-7}N\times x^2=9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}$

$x^2=\frac{9.0\times 10^9 Nm^2C^{-2}\times 6.7^2\times 10^{-18} C^2}{8.2\times 10^{-7}N}$

$x=\frac{3\times 6.7}{\sqrt{820} }m$

$x=.701m$

therefore distance between two given charges is =.701m