Question

A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength equal to 5.8 × 105 newtons/coulomb?
(k = 9.0 × 109 newton·meter2/coulomb2)

0.2 × 10-2 meters
4.8 × 10-3 meters
3.0 × 10-2 meters
5.8 × 105 meters

Please help I keep failing this quiz. -_-

Answer 1

Answer:

Distance, $d=4.8\times 10^{-3}\ m$

Explanation:

It is given that,

Electric charge on a ball, $q=1.5\times 10^{-9}\ C$

Electric field strength, $E=5.8\times 10^5\ N/C$

The electric field strength is given by :

$E=k\dfrac{q}{d^2}$

$r=\sqrt{\dfrac{kq}{E}}$

$r=\sqrt{\dfrac{9\times 10^9\times 1.5\times 10^{-9}}{5.8\times 10^5}}$

r = 0.0048 meters

or

$r=4.8\times 10^{-3}\ m$

Hence, the correct option is (b).

Answer 2

E=Fe/q                                                      
5.8x10^5N/C=Fe/(1.5x10^-9C)             
Fe=(5.8x10^5N/C)(1.5x10^-9C)
Fe=8.7x10^-4N

Fe=kq1q2/r²
8.7X10^-4N=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/r²
r²=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/(8.7x10^-4N)
√r²=√0.00002325
The final answer is r=4.8x10-3m