Answer to Question #297693 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.   

 

3. 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.

4. 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

1)

$\begin{vmatrix} E \\ \end{vmatrix}=\frac{KQ}{r^2}$

$=\frac{9×10^9×4×10^{-9}}{2^2}=9N/C$

2)

$\begin{vmatrix} E \\ \end{vmatrix}$ $=\frac{F}{q}$

$=\frac{2.75×10^{-4}}{60×10^{-9}}=4583.3N/C$

3)

a)

$\begin{vmatrix} F \\ \end{vmatrix}=\frac{Kq_1q_2}{r^2}$

$=\frac{9×10^9×43×17×(10^{-9})^2}{(6×10^{-2})^2}$

$=1.8275×10^{-3}N$

b)

$\begin{vmatrix} E_{q_1} \\ \end{vmatrix}=\frac{F}{q}=\frac{1.8275×10^{-3}}{17×10^{-9}}$

$=1.075×10^5N/C$

$\begin{vmatrix} E_{q_2} \\ \end{vmatrix}=\frac{1.8275×10^{-3}}{43×10^{-9}}=4.25×10^4N/C$

4)

$E=\frac{KQ}{r^2}=\frac{9×10^9×5×10^{-8}}{(3×10^{-2})^2}=5×10^5N/C$

$V=\frac{KQ}{r}=\frac{9×10^9×5×10^{-8}}{(3×10^{-2})}=1.5×10^4V$