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.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

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