Respuesta :
Answer:
(a) 1745.36 V
(b) 8312.5 N/C
Explanation:
q1 = 8.6 nC at origin
q2 = 5.3 nC at x = 10 cm
(a)
Let the electric potential at A is V.
The formula for the electric potential is given by
[tex]V=\frac{Kq}{r}[/tex]
The electric potential at A due to charge q1 is
[tex]V_{1}=\frac{Kq_{1}}{r_{1}}[/tex]
[tex]V_{1}=\frac{9\times 10^{9}\times 8.6\times 10^{-9}}{0.14}[/tex]
V1 = 552.86 V
the electric potential at A due to charge q2 is
[tex]V_{2}=\frac{Kq_{2}}{r_{2}}[/tex]
[tex]V_{2}=\frac{9\times 10^{9}\times 5.3\times 10^{-9}}{0.04}[/tex]
V2 = 1192.5 V
The total potential at A is
V = V1 + V2 = 552.86 + 1192.5 = 1745.36 V
(b)
The formula for the electric field is given by
[tex]E = \frac{Kq}{r^{2}}[/tex]
The electric field due to charge q1 at B is
[tex]E_{1} = \frac{Kq_{1}}{r_{1}^{2}}[/tex] towards right
[tex]E_{1} = \frac{9\times 10^{9}\times 8.6\times 10^{-9}}{0.06^{2}}[/tex]
E1 = 21500 N/C towards right
The electric filed due to charge q2 at B is
[tex]E_{2} = \frac{Kq_{2}}{r_{2}^{2}}[/tex] towards left
[tex]E_{2} = \frac{9\times 10^{9}\times 5.3 \times 10^{-9}}{0.04^{2}}[/tex]
E2 = 29812.5 N/C towards left
ne electric filed at B is
E = E2 - E1
E = 29812.5 - 21500 = 8312.5 N/C towards origin
