Each 2.0-m length of a long cylinder (radius = 4.0 mm) has a charge of 4.0 nC distributed uniformly throughout its volume. What is the magnitude of the electric field at a point 5.0 mm from the axis of the cylinder?

5618N/C

Explanation:

According to the question, for every 2m length of the cylinder, there is a total charge of 4.0nC (an equivalent of 4×10^(-9)C).

Therefore by comparison, for a 5mm (0.005m) length,

The quantity of charge distributed through out its volume would be

Quantity of charge(Q) =

(4nC × 0.005m)÷(2m)

The quantity of charge (Q) at this length is then 0.01nC or 1 × 10^(-11)C.

To evaluate the magnitude of electric field at a point, the formulae used is provided below:

E= kQ÷r²

However, in this case this in converted into an integral and evaluated over the radius of the cylinder.

Therefore,

E= (8.99×10^(9) Nm²/C²× 1×10^(-11)C) ÷ (0.004²)

Since radius(r)=4mm=0.004m.

E= 5618.75N/C.

a.   mass number from atomic number

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2