What is the density of ocean water if 308.19g fills a 300mL container?
1 answer:
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Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m
Answer:
Current, I = 0.0011 A
Explanation:
It is given that,
Diameter of rod, d = 2.56 cm
Radius of rod, r = 1.28 cm = 0.0128 m
The resistivity of the pure silicon,
Length of rod, l = 20 cm = 0.2 m
Voltage,
The resistivity of the rod is given by :
R = 893692.30 ohms
Current flowing in the rod is calculated using Ohm's law as :
V = I R
I = 0.0011 A
So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.