Can someone help me this question
1 answer:
You might be interested in
Answer:
a)
b)
c)
Explanation:
Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:
Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.
When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.
a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:
b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.
The new surface charge density can be calculated as follows:
c) The electric field outside the cylinder can be found by Gauss' Law:
We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.