Answer to Question #308382 in Electricity and Magnetism for Hotsog

Question

1. 2 charged particles reacted with a force of 0.31 N with a distance of 1.5 cm, the 2nd charge contains 54 nC. Find the value of the first charge.

2. Using eq. 2, solve for the Force of 2 charges, q1 = -63 nC and q2 = +42 nC with a distance of 5.5 cm. (Note: ϵ0 has a value of 8.85 x 10⁻¹² F/m) (F= N m/ C²)

3. Find the charge of a test charge with an electric field of 6.0 x 10⁻⁶ N/C and a force of 0.25 N

4. Solve for the Electric field of a point charge q, with a value of 2.9 nC and a distance of of 2.4 m

5. Find the value of the point charge with an electric field of 12 N/C and a distance 3.3 m

Answer 1

(1)

$F=\frac{kq_1q_2}{r^2}$

$$q_1=\frac{Fr^2}{kq_2}=\frac{0.31\times0.015^2}{9\times10^9\times54\times10^{-9}}=1.44\times10^{-7}N$$

(2) force

$F=\frac{kq_1q_2}{r^2}$

$$F=-\frac{9\times10^9\times63\times10^{-9}\times42\times10^{-9}}{0.055^2}=-7.88\times10^{-3}N$$

(3) charge

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

$q=\frac{0.25}{6.0\times10^{-6}}=41.66\times10^{+3}C$

(4)

$E=\frac{kq}{r^2}$

$E=\frac{9\times10^9\times2.9\times10^{-9}}{2.4^2}=4.54N/C$

(5)

$E=\frac{kq}{r^2}$

$q=\frac{Er^2}{k}$

$q=\frac{12\times3.3^2}{9\times10^9}=14.52\times10^{-9}C$