Answer to Question #297691 in Electric Circuits for HYEHYE

Question

1. What is the magnitude of the electric field at a field point 2 m from a point charge q₀ = 4 nC?  
2. Determine the electric field of a –60-nC charge that is experiencing an electric force of

2.75 x 10^-4 N.

1. Suppose two point charges, q₁ = + 43 nC and q₂ = –17 nC, are separated by a distance of 6 cm. Determine the:

(a) electric force interacting between the two charges; and

(b) electric field of each charge.

1. Determine the electric field strength and potential in air at a distance of 3 cm from a charge of 5 X 10^-8 C

Answer 1

(a)

$E=\frac{kq}{r^2}=\frac{9\times10^9\times4\times10^{-9}}{2^2}=9N/C$

$(2)$

$E=\frac{F}{q}$

$E=-\frac{2.75\times10^{-4}}{60\times10^{-9}}=-4583.33N/C$C

$$F=\frac{kq_1q_2}{r^2}=-\frac{9\times10^9\times43\times10^{-9}\times17\times10^{-9}}{0.06^2}=-1.83\times10^{-3}N$$

Electric field

$$E_1=\frac{kq_1}{r^2}=\frac{9\times10^9\times43\times10^{-9}}{0.03^2}=430KN/C$$

$$E_2=\frac{kq_2}{r^2}=-\frac{9\times10^9\times17\times10^{-9}}{0.03^2}=-170KN/C$$

(b)

Electric field

$$E=\frac{kq}{r^2}=-\frac{9\times10^{9}\times5\times10^{-9}}{0.03^2}=-50KN/C$$

Potential

$$V=\frac{Kq}{r}=-\frac{9\times10^9\times5\times10^{-9}}{0.03}=1500KV$$