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Example · Example 5

Q.Starting from Coulomb's law, define the electric field at a point and derive the expression for the electric field due to an isolated point charge qq at distance rr. Hence find the magnitude of the field 0.1 m0.1\ \text{m} from a charge of +2×10−9 C+2\times10^{-9}\ \text{C}.

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The electric field at a point is defined as the force per unit positive test charge, in the limit that the test charge is small enough not to disturb the source: E⃗=lim⁡q0→0F⃗/q0\vec{E}=\lim_{q_0\to0}\vec{F}/q_0.

Since the Coulomb force on a test charge q0q_0 at distance rr from a source charge qq is F=kqq0/r2F=kqq_0/r^2, dividing by q0q_0 gives

E=kqr2E = \frac{kq}{r^2}

directed radially outward from a positive source (radially inward for a negative source). …

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