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The MIDLAND INSURANCE COMPANY, a Corporation
v.
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No. 85-C-1537.
Supreme Court of Louisiana.
February 25, 1986.
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*684 George D. Aaron, Dale M. Steele, Aaron & Steele, New Orleans, for plaintiff-applicant.
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DENNIS, Justice.
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// Boost string_algo library constants.hpp header file —————————//

//
// (See accompanying file LICENSE_1_0.txt or copy at
//

// See for updates, documentation, and revision history.

#ifndef BOOST_STRING_CONSTANTS_HPP
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} // BOOST_ALGORITHM_NAMESPACE

}

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Why does the electric field of a point charge appear to have a radial component?

In this question it was mentioned that the electric field of a point charge

tends to point towards the charge and away from the centre of the
charge.

Is there a reason why the electric field has a radial component?

A:

Your question is not adequately framed. What are we talking about? “Why does a point charge have a radial component in its electric field?” No one knows. Likewise, a sphere has a uniform electric field, although it clearly has no electric field points. A charged dielectric material has a uniform electric field, but again, it’s hard to say whether its field is all radial.
What do we know about the electric field of a point charge? We know that the electric field varies with distance from the charge and that it does so in a radial fashion. As a point charge approaches the observer (or the grounded surface) the electric field points toward it (positive charge “pushes” the electric field toward itself) and is strongest at the point closest to the charge.
Why do we know that the electric field varies with distance from the charge and is radial? Because we know the electric field of the positive charge, $E^{+}(r)$, and we know that the electric field of a negative charge, $E^{ – }(r)$, is such that the magnitudes of the positive field and the negative field add up to zero at all points. Thus, the electric field at any point is always equal to the electric field of the total charge divided by the amount of positive charge in the region. That is, the electric field of a point charge equals $$E(r) = \frac{Q}{4\pi\epsilon_0 r^2} \text{ }\Leftrightarrow\text{ }E(r) = \frac{q(r)}{4\pi\epsilon_0 r}$$ The

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