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UkoKoshka [18]
3 years ago
5

At what point will the electric field of a charged object be strongest?

Physics
2 answers:
Lelu [443]3 years ago
6 0

Explanation:

The expression from electric field is as follows;

E=\frac{kq}{r^{2} }

Here, E is the electric field, k is constant, r is the distance and [tex]q is the charges.

The electric field is inversely proportional to the distance. It is directly proportional to the charge.

As the distance from the source charge increases the electric field also gets decreased. The electric field gets increased if the distance from the source decreases.

The electric field lines are the surrounding around the charges in which the force due to the electric field can be experienced.

The electric field is strongest near the charged object.

Alexandra [31]3 years ago
4 0

The electric field of a charged object be strongest at the point where the electric field lines are closest together.

To add, the electric field is <span>a region around a charged particle or object within which a force would be exerted on other charged particles or objects.</span>

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A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per m
Natali [406]

(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

<h3>Electric field at a point on the Gaussian surface</h3>

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

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<h3>At a distance of 5.5 m</h3>

E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C

<h3>At a distance of 2.5 m</h3>

E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

Learn more about electric field here: brainly.com/question/14372859

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