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kifflom [539]
3 years ago
11

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 eq

ual to 5.8 × 105 newtons/coulomb?
Physics
1 answer:
Veronika [31]3 years ago
3 0
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 -  two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3 
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