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What is the formula for the force due to a linear charge distribution?
Although charges are quantised we can consider a large collec-tion of charges in a region as a continuous distribution. The con-tinuous distribution of charges may be of three types—
- one-dimensional,
- two-dimensional and
- three-dimensional. These distributions of charges are called linear charge distribu-tion, surface charge distribution and volume charge distribu-tion, respectively.
i) Linear charge distribution : Let there be a continuous distribution of charge along a straight line or a curved line placed in vacuum or in air. The linear charge density = \(\left(\frac{\text { charge }}{\text { length }}\right)\) = λ. Suppose, dl be an infinitesimally small element of this line charge distribution [Fig.]. Charge on the element dl is dq = λdl.
Suppose, a test point charge q0 is at a distance r from the small element dl.
Now, the force acting on q0 due to the charge dq is given by,
\(d \vec{F}\) = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 d q}{r^2} \hat{r}\) ; where \(\hat{r}\) = \(\frac{\vec{r}}{r}\) = unit vector along \(\vec{r}\)
= \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 \lambda d l}{r^2} \hat{r}\)
∴ The total force acting on q0 due to the whole line charge is given by,
\(\vec{F}\) = \(\frac{q_0}{4 \pi \epsilon_0} \int_l \frac{\lambda}{r^2} \hat{r} d l\) ……… (1)
l indicates the line along which the integration extends.
In CGS system the above equation is (replacing ε0 by \(\frac{1}{4 \pi}\)),
\(\vec{F}\) = \(q_0 \int_l \frac{\lambda}{r^2} \hat{r} d l\) …… (2)
ii) Surface charge distribution: Let there be a continuous dis-tribution of charge over a surface placed in vacuum or in air. The surface charge density = \(\left(\frac{\text { charge }}{\text { area }}\right)\) = σ.
Suppose, ds be an infinitesimally small surface element of this surface distribution of charge [Fig.]. Charge on the surface element ds is dq = σds.
Suppose, a test point charge q0 is at a distance r from the small surface element ds.
Now, the force acting on q0 due to the charge dq is given by,
\(d \vec{F}\) = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 d q}{r^2} \hat{r}\) = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 \sigma d s}{r^2} \hat{r}\)
∴ The total force acting on q0 due to the whole surface charge distribution is given by,
\(\vec{F}\) = \(\frac{q_0}{4 \pi \epsilon_0} \int_s \frac{\sigma}{r^2} \hat{r} d s\) ….. (3)
s indicates the surface over which the integration extends. In CGS system the above equation is,
\(\vec{F}\) = \(q_0 \int_s \frac{\sigma}{r^2} \hat{r} d s\) …… (4)
iii) Volume or spatial charge distribution: Let there be a con-tinuous distribution of charge over a volume (such as a cube or a sphere) placed in vacuum or in air. The volume density of charge = \(\left(\frac{\text { charge }}{\text { volume }}\right)\) = ρ.
Suppose, dv be an infinitesimally small volume element of this volume distribution of charge [Fig.]. Charge on the volume element dv is dq = ρdv.
Suppose, a test point charge q0 is at a distance r from the small volume element dv.
Now, the force acting on q0 due to the charge dq is given by,
\(d \vec{F}\) = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 d q}{r^2} \hat{r}\) = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0 \rho d v}{r^2} \hat{r}\)
∴ The total force acting on q0 due to the whole volume charge distribution is given by,
\(\vec{F}\) = \(\frac{q_0}{4 \pi \epsilon_0} \int_v \frac{\rho}{r^2} \hat{r} d v\) …… (5)
v indicates the volume to which the integration extends.
In CGS system the above equation is,
\(\vec{F}\) = \(q_0 \int_v \frac{\rho}{r^2} \hat{r} d v\) ………. (6)