Respuesta :
Answer: 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.
