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Class XII problemVariation of electric field and electric potential with distance from a point charge.

Answer»

Electrical force on a point unit CHARGE which is at a distance x, exerted by a point charge q = electric field =
\vec{E}=\frac{1}{4 \pi \epsilon}* \frac{q}{x^2}
the direction of electric field is ALONG the line joining the charges.

variation of electric field with distance from the point charge =
\frac{d\vec{E}}{dx}=the\ derivative=\frac{1}{4\pi \epsilon}\frac{(-2)q}{x^3}}\\\\=-\frac{q}{4\pi \epsilon\ x^3}

The electric potential at point P (at distance x from origin) due to the electric point  charge q at the origin = V
     V = energy REQUIRED to move a unit charge q₀ from infinity, against the electric force due to charge q, to the point P at a distance x away from q.

V=U/q_0=\int\limits^x_\infty {\vec{F}.} \, \vec{ds} =\int\limits^x_\infty {F} \, dx\\\\=\int\limits^x_\infty {\frac{1}{4 \pi \epsilon}\frac{q}{x^2}} \, dx \\\\V=-\frac{1}{4 \pi \epsilon}*\frac{q}{x}

the variation of electric potential with distance is 
\frac{dV}{dx}=\frac{1}{4 \pi \epsilon}\frac{q}{x^2}=\vec{E}


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