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III. Long Answer Questions · Q6

Q.Derive an expression for electrostatic potential due to a point charge.

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Step 1. Setup. A positive charge qq sits at the origin; a unit positive test charge is brought from infinity (where V=0V=0) to a point P at distance rr.

Step 2. Work integral. The work done against the repulsive Coulomb force over this path is W=∫∞r−F⃗⋅dl⃗W=\displaystyle\int_{\infty}^{r} -\vec{F}\cdot d\vec{l}; carrying out this integration using F=kq/x2F=kq/x^2 along the radial path gives W=kq/rW=kq/r.

Step 3. Result. Since V=WV=W per unit test charge, V(r)=14πε0qrV(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}: positive and decreasing with rr around a positive source, negative and increasing (toward zero) with rr around a negative source. …

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