Respuesta :
Answer:
[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]
Explanation:
Given:
- one charge, [tex]Q=2\ C[/tex]
- another charge, [tex]q=10^{-9}\ C[/tex]
- distance between the two charges, [tex]r=1\ m[/tex]
- force between the charges, [tex]F=18\ N[/tex]
We know from the Coulomb's law:
[tex]F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}[/tex] ...........(1)
where:
[tex]\epsilon_0=[/tex] permittivity of free space
Also we have electric field due to Q at q (which is at a distance of 1 m):
[tex]E=\frac{1}{4\pi.\epsilon_0} \times \frac{Q}{r^2}\ [N.C^{-1}][/tex] ...........(2)
From (1) & (2)
[tex]E=\frac{F}{q}[/tex]
[tex]E=\frac{18}{10^{-9}}[/tex]
[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]
Answer:
magnitude of the electric field = 1.8 × [tex]10^{10}[/tex] N/C
Explanation:
given data
charge Q =2 C
charge q = 1 nC = 1 × [tex]10^{-9}[/tex] C
distance r = 1 m
electric force = 18 N
solution
we know that here electric field due to Q is
E = k × [tex]\frac{Q}{r^2}[/tex] ..............1
here q distance is 1 m
now we apply here Coulomb’s law and get here electric force
F = k ×[tex]\frac{Q*q}{r^2}[/tex] ..........2
we know constant k = 8.988 × [tex]10^{9}[/tex] Nm²/C²
so from above both equation we get
electric filed = [tex]\frac{force}{charge}[/tex] .................3
put here value
electric filed = [tex]\frac{18}{1*10^{-9}}[/tex] = 1.8 × [tex]10^{10}[/tex] N/C
