Respuesta :
Answer:
The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
Explanation:
Given that,
Charge = 3 μC
Radius a=1 m
Distance = 5 m
We need to calculate the electric field at any point on the axis of a charged ring
Using formula of electric field
[tex]E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
[tex]E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{1}=1.0183\times10^{3}\ N/C[/tex]
Using formula of electric field again
[tex]E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{2}=-1.064\times10^{3}\ N/C[/tex]
We need to calculate the resultant electric field
Using formula of electric field
[tex]E=E_{1}+E_{2}[/tex]
Put the value into the formula
[tex]E=1.0183\times10^{3}-1.064\times10^{3}[/tex]
[tex]E=-0.045\times10^{3}\ N/C[/tex]
We need to calculate the force exerted on an electron
Using formula of electric field
[tex]E = \dfrac{F}{q}[/tex]
[tex]F=E\times q[/tex]
Put the value into the formula
[tex]F=-0.045\times10^{3}\times(-1.6\times10^{-19})[/tex]
[tex]F=7.2\times10^{-18}\ N[/tex]
Hence, The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
