Respuesta :
Answer:
The smaller sphere has higher value of electric field than the threshold value of electric field.
Solution:
As per the question:
The radius of the first metal sphere, R = 0.1 m
The radius of the first metal sphere, R' = 0.01 m
Charge on the sphere, Q = [tex]1.0\tiems 10^{- 6}\ C[/tex]
The value of the critical electric field, [tex]\vec{E_{c}} = 3\times 10^{6} V/m[/tex]
Now,
The electric field on the surface of a sphere is given by:
[tex]\vec{E} = K\frac{Q}{R^{2}}[/tex]
where
K = Electrostatic constant = [tex]\frac{1}{4\pi\epsilon_{o}} = 9\times 10^{9}N.m^{2}C^{- 2}[/tex]
Now, for the first metal sphere:
[tex]\vec{E} = K\frac{Q}{R^{2}}[/tex]
[tex]\vec{E} = (9\times 10^{9})\times \frac{(1.0\tiems 10^{- 6})}{0.1^{2}} = 9\times 10^{5}\ V/m[/tex]
Now, for the second metal sphere:
[tex]\vec{E'} = K\frac{Q}{R'}[/tex]
[tex]\vec{E'} = (9\times 10^{9})\times \frac{(1.0\tiems 10^{- 6})}{0.01^{2}} = 9\times 10^{7}\ V/m[/tex]
The Electric field for second metal sphere,i.e., smaller sphere is higher than the threshold or critical value for the Electric field thus smaller sphere discharges quickly as above critical field breakdown occurs.
