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rodikova [14]
3 years ago
11

What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?

Physics
1 answer:
Debora [2.8K]3 years ago
6 0
Thinking the small glass bead as a single point charge, the electric field generated by it is given by
E(r) = k_e  \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant
Q=8.0 nC=8.0 \cdot 10^{-9} C is the charge of the bead
r=3.0 cm=0.03 m is the distance at which we calculate the field.

Using these data, we find:
E=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(8.0 \cdot 10^{-9} C}{0.03 m)^2}=8.0 \cdot 10^4 N/C
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I hope it helps you ;)
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