Physics > Electrostatics > 6.0 Electric field
Electrostatics
1.0 Introduction
2.0 Electric charge
3.0 Coulomb's law
3.1 Coulomb's law in vector relations
3.2 Comparision between coulomb's force and gravitational force
4.0 Principle of superposition
5.0 Continuous charge distribution
6.0 Electric field
6.1 Electric field due to a point charge
6.2 Electric field due to a ring of charge
6.3 Electric field due to a line of charge
7.0 Electric field lines
8.0 Insulators and conductors
9.0 Gauss's law
9.1 Electric field due to a point charge
9.2 Electric field due to a linear charge distribution
9.3 Electric field due to a plane sheet of charge
9.4 Electric field near a charged conducting surface
9.5 Electric field due to a charged spherical shell or solid conducting surface
9.6 Electric field due to a solid sphere of charge
10.0 Work done
10.1 Work done by electrical force
10.2 Work done by external force
10.3 Relation between work done by electrical & external force
11.0 Electric potential energy
12.0 Electric Potential
12.1 Properties
12.2 Use of Potential
12.3 Potential Due to Point Charge
12.4 Potential due to a Ring
12.5 Potential Due to Uniformly charged Disc
12.6 Potential Due To Uniformly Charged Spherical Shell
12.7 Potential Due to Uniformly Charged Solid Sphere
13.0 Electric dipole
13.1 Electric field due to a dipole at axial point
13.2 Electric field on equatorial line
13.3 Electric field at any point
13.4 Dipole in an external electric field
13.5 Potential due to an electric dipole
6.3 Electric field due to a line of charge
3.2 Comparision between coulomb's force and gravitational force
6.2 Electric field due to a ring of charge
6.3 Electric field due to a line of charge
9.2 Electric field due to a linear charge distribution
9.3 Electric field due to a plane sheet of charge
9.4 Electric field near a charged conducting surface
9.5 Electric field due to a charged spherical shell or solid conducting surface
9.6 Electric field due to a solid sphere of charge
10.2 Work done by external force
10.3 Relation between work done by electrical & external force
12.2 Use of Potential
12.3 Potential Due to Point Charge
12.4 Potential due to a Ring
12.5 Potential Due to Uniformly charged Disc
12.6 Potential Due To Uniformly Charged Spherical Shell
12.7 Potential Due to Uniformly Charged Solid Sphere
13.2 Electric field on equatorial line
13.3 Electric field at any point
13.4 Dipole in an external electric field
13.5 Potential due to an electric dipole
Consider a positive charge $q$ is distributed uniformly along a line with length $2a$, lying along the $y-$ axis between $y=-a$ and $y=+a$ as shown in the figure.
Charge per unit length is, $\lambda = \frac{q}{{2a}} \quad ...(i)$
The electric field due to an infinitesimally small section $dl$ having charge $dq$ at any point $P$ which is at a distance $x$ from the center of the ring is given by, $$d\overrightarrow E = \frac{{kdq}}{{{r^2}}}$$
Resolving electric field into a horizontal and vertical component.
$$d\overrightarrow E = d{\overrightarrow E _x} + d{\overrightarrow E _y}$$
Integrating both sides, $$\int {d\overrightarrow E } = \int {d{{\overrightarrow E }_x}} + \int {d{{\overrightarrow E }_y}} $$
If we consider top and bottom segments of the line, we see that the contributions of the field at point $P$ from these segments have same $x-$ component but opposite $y-$ components. Hence, the total $y-$ component of field due to this pair of segments is zero.
$$d\overrightarrow E = d{\overrightarrow E _x} + d{\overrightarrow E _y}$$
Integrating both sides, $$\int {d\overrightarrow E } = \int {d{{\overrightarrow E }_x}} + \int {d{{\overrightarrow E }_y}} $$
If we consider top and bottom segments of the line, we see that the contributions of the field at point $P$ from these segments have same $x-$ component but opposite $y-$ components. Hence, the total $y-$ component of field due to this pair of segments is zero.
$${\int {d{{\overrightarrow E }_y}} = 0}$$ So, $$\int {d\overrightarrow E } = \int {d{{\overrightarrow E }_x}} $$ As $d{E_x} = dE\cos \theta $, $$\int {dE} = \int {\frac{{kdq}}{{{r^2}}}\cos \theta } $$ Also, $\cos \theta = \frac{x}{r} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, $$\int {dE} = \int {\frac{{kxdq}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$$$\int {dE} = \int {\frac{{kx\left( {\lambda dy} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} \quad ...(ii)$$
From equation $(i)$ and $(ii)$ we get, $$\int {dE} = \int {\frac{{kxqdy}}{{2a{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$$$\int {dE} = \frac{{kq}}{{2a}}\int {\frac{{xdy}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$
From equation $(i)$ and $(ii)$ we get, $$\int {dE} = \int {\frac{{kxqdy}}{{2a{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$$$\int {dE} = \frac{{kq}}{{2a}}\int {\frac{{xdy}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$
Integration using proper limits, $$\int\limits_0^E {dE} = \frac{{kq}}{{2a}}\int\limits_{ - a}^{ + a} {\frac{{xdy}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} $$$$\left[ E \right]_0^E = \frac{{kq}}{{2ax}}\left[ {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right]_{ - a}^{ + a}$$$$(E - 0) = \frac{{kq}}{{2ax}}\left[ {\frac{a}{{\sqrt {{x^2} + {a^2}} }} - \left( {\frac{{ - a}}{{\sqrt {{x^2} + {a^2}} }}} \right)} \right]$$$$E = \frac{{kq}}{{2ax}}\left( {\frac{{2a}}{{\sqrt {{x^2} + {a^2}} }}} \right)$$ or $$E = {E_x} = \frac{{kq}}{{x\sqrt {{x^2} + {a^2}} }} = \frac{q}{{4\pi {\varepsilon _0}x\sqrt {{x^2} + {a^2}} }}$$ and $${E_y} = 0$$
So, the electric field is only along $x-$ axis.
Note:
- If $x > > a$, So $\sqrt {{x^2} + {a^2}} \approx x$. Therefore, $${E_x} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{x^2}}}$$
If point $P$ is very far from the line charge, then the field at $P$ is same as that of a point charge. - If we make the line of charge longer & longer. Also adding charge in proportion to the total length so that $\lambda $ (charge per unit length) remains constant. Mathematically, $$E = \frac{q}{{4\pi {\varepsilon _0}x\sqrt {{x^2} + {a^2}} }}$$
Divinding numerator and denominator by $2a$ we get, $$E = \frac{{\frac{q}{{2a}}}}{{4\pi {\varepsilon _0}x\frac{{\sqrt {{x^2} + {a^2}} }}{{2a}}}}$$$$E = \frac{\lambda }{{2\pi {\varepsilon _0}x\sqrt {\frac{{{x^2}}}{{{a^2}}} + 1} }}$$ As $a > > x$. So $a > > x$,
$$E = \frac{\lambda }{{2\pi {\varepsilon _0}x}}$$ or $$E \propto \frac{1}{x}$$
So, the magnitude of electric field depends only on distance of point $P$ from the infinite line of charge.
