Respuesta :
Answer:
charge will be [tex]0.123\times 10^{-9}C.[/tex]
Explanation:
We have given that electric field due to point charge is E=2.4 N/C
Distance from point charge R = 68 cm = 0.68 m
We know that electric field due to point charge is given as [tex]E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^2}=\frac{KQ}{R^2}[/tex]
So [tex]2.4=\frac{9\times 10^9Q}{0.68^2}[/tex]
[tex]Q=0.123\times 10^{-9}C[/tex]
So charge will be [tex]0.123\times 10^{-9}C.[/tex]
Answer:
Charge, [tex]q=1.23\times 10^{-10}\ C[/tex]
Explanation:
It is given that,
Electric field due to a point charge, E = 2.4 N/C
Distance from the point charge, d = 68 cm = 0.68 m
The formula to find the electric field for a point charge at a distance d is given by :
[tex]E=\dfrac{kq}{d^2}[/tex]
[tex]q=\dfrac{Ed^2}{k}[/tex]
[tex]q=\dfrac{2.4\times (0.68)^2}{9\times 10^9}[/tex]
[tex]q=1.23\times 10^{-10}\ C[/tex]
So, the magnitude of a point charge is [tex]1.23\times 10^{-10}\ C[/tex]. Hence, this is the required solution.
