Respuesta :
Answer:
a)[tex]E=50.53\times 10^{6}\ N/C[/tex]
The direction will be negative direction.
b)[tex]E=268.22\times 10^{6}\ N/C[/tex]
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
[tex]E=K\dfrac{q}{r^2}[/tex]
[tex]E_1=K\dfrac{q_1}{r_1^2}[/tex]
[tex]E_2=K\dfrac{q_2}{r_2^2}[/tex]
Now by putting the va;ues
a)
[tex]E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C[/tex]
[tex]E_1=72.11\times 10^{6}\ N/C[/tex]
[tex]E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C[/tex]
[tex]E_2=21.58\times 10^{6}\ N/C[/tex]
The net electric field
[tex]E=E_1-E_2[/tex]
[tex]E=50.53\times 10^{6}\ N/C[/tex]
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
[tex]E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C[/tex]
[tex]E_1=120.3\times 10^{6}\ N/C[/tex]
[tex]E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C[/tex]
[tex]E_2=147.92\times 10^{6}\ N/C[/tex]
The net electric field
[tex]E=E_1+E_2[/tex]
[tex]E=268.22\times 10^{6}\ N/C[/tex]
The direction will be positive direction.
