Respuesta :
Answer:
The electric field at a point 60.0 cm from the centre is [tex]E=1.55\times10^5\ \rm N/C[/tex]
Explanation:
Given:
- Radius of the spherical cavity=7.00 cm
- Radius of the metal sphere = 17.5 cm
- magnitude of the point charge Q=[tex]6.2\ \rm \mu C[/tex]
Since the solid sphere is Metallic and we know that there should be no charge inside a metal due to property. So the Equal magnitude but opposite in sign of the charge will be at that inner surface of the cavity and finally to the outer surface of the metal sphere.
Let E be the electric field due to this at the 60 cm from the centre given by
[tex]E=\dfrac{KQ}{r^2}\\E=\dfrac{9\times10^9\times6.2\times10^{-6}}{0.6^2}\\\\E=1.55\times10^5\ \rm N/C[/tex]
The direction of electric Field is radially outwards.
The value of electric field at a point 60.0 cm from the center of a metal sphere is 1.55×10⁵ N/C.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge. It can be given as,
[tex]E=k\dfrac{Q}{r^2}[/tex]
Here, (Q) is the charge, (r) is the radius and (k) is the coulombs constant ([tex]9\times10^9\rm Nm^2/C^2[/tex]).
A spherical cavity of radius 7.00 cm is at the center of a metal sphere of radius 17.5 cm . A point charge 6.20 μC rests at the very center of the cavity, whereas the metal conductor carries no net charge.
The electric field at a point 60.0 cm or 0.6 m from the center has to be found out. As, the value of electric charge is 6.20 μC . Thus, put the values in the above formula as,
[tex]E=(9\times10^{9})\dfrac{6.2\times10^{-6}}{(0.6)^2}\\\E=1.55\times10^{5}\rm N/C[/tex]
Thus, the value of electric field at a point 60.0 cm from the center of a metal sphere is 1.55×10⁵ N/C.
Learn more about electric field here;
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