Answer :
Electrical force on a point unit charge which is at a distance x, exerted by a point charge q = electric field =
[tex]\vec{E}=\frac{1}{4 \pi \epsilon}* \frac{q}{x^2}[/tex]
the direction of electric field is along the line joining the charges.
variation of electric field with distance from the point charge =
[tex]\frac{d\vec{E}}{dx}=the\ derivative=\frac{1}{4\pi \epsilon}\frac{(-2)q}{x^3}}\\\\=-\frac{q}{4\pi \epsilon\ x^3}[/tex]
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.
[tex]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}[/tex]
the variation of electric potential with distance is
[tex]\frac{dV}{dx}=\frac{1}{4 \pi \epsilon}\frac{q}{x^2}=\vec{E}[/tex]
[tex]\vec{E}=\frac{1}{4 \pi \epsilon}* \frac{q}{x^2}[/tex]
the direction of electric field is along the line joining the charges.
variation of electric field with distance from the point charge =
[tex]\frac{d\vec{E}}{dx}=the\ derivative=\frac{1}{4\pi \epsilon}\frac{(-2)q}{x^3}}\\\\=-\frac{q}{4\pi \epsilon\ x^3}[/tex]
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.
[tex]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}[/tex]
the variation of electric potential with distance is
[tex]\frac{dV}{dx}=\frac{1}{4 \pi \epsilon}\frac{q}{x^2}=\vec{E}[/tex]