Duas cargas elétricas Q¹= 15.10-seis ao quadrado C e Q² = 150.10-seis ao quadrado, estão separadas por uma distância de 0,1m, no vácuo. Determine a intensidade da força elétrica…

Question

asked Sep 28, 2021 65.5k views

1 Answer

Paolo · Answer 1 · 2021-10-02T18:55:17+0000

Answer:

The electric field intensity of electric force is
$1.35*10^(7)\ N/C$

Step-by-step explanation:

Given that,

First charge
$q_(1)=15*10^(-6)\ C$

Second charge
$q_(2)=150*10^(-6)\ C$

Distance = 0.1 m

We need to calculate the electric field intensity of electric force

Using formula of intensity of electric force

E=(F)/(q_(2))

E=(kq_(1))/(r^2)

Put the value into the formula

E=(9*10^(9)*15*10^(-6))/((0.1)^2)

$E=13500000\ N/C$

$E=1.35*10^(7)\ N/C$

Hence, The electric field intensity of electric force is
$1.35*10^(7)\ N/C$