Question

A wire of length L is placed along x - axis with one end at the origin. The linear charge density of the wire varies with distance x from the origin as \[\lambda {\text{ }} = \;{\lambda _0}\;\frac{{{x^3}}}{L}\], where $λ_0$ is a positive constant. The total charge Q on the rod is
A. $Q = \frac{{{\lambda _0}{L^2}}}{3}$
B. $Q = \frac{{{\lambda _0}{L^3}}}{3}$
C. $Q = \frac{{{\lambda _0}{L^3}}}{4}$
D. $Q = \frac{{{\lambda _0}{L^4}}}{5}$

Answer 1

Hint: Linear charge density λ is defined as amount of charge per unit length. Its S.I. unit is coulomb per metre.
\[\lambda = \frac{{dq}}{{dl}}\]

Complete answer:
consider a small portion of a wire of length \[dl\]with a charge over it as\[dq\]. Let λ0 be the initial linear charge density.
Therefore using the formula, $dq = \lambda \;dl$
As \[\lambda {\text{ }} = \;{\lambda _0}\;\left( {{x^3}/L} \right)\]
Therefore,
$dq = {\lambda _0}\frac{{{x^3}}}{L}\;dx$

Integrating both sides to find the linear charge density.
\[\int\limits_0^Q {dq} = {\lambda _0}\int\limits_0^L {\frac{{{x^3}}}{L}\;dx} \] (\[{\lambda _0}\]is given to be a positive constant)
$Q = \frac{{{\lambda _0}}}{L}\left| {\frac{{{x^4}}}{4}} \right|_0^L$
$Q = \frac{{{\lambda _0}}}{L}\frac{{{L^4}}}{4}$

On cancelling L present in denominator we get,
$Q = \frac{{{\lambda _0}{L^3}}}{4}$

Thus, the correct answer is C.

Note: here are various different kinds of charge densities. Linear charge density is one of the. Others are surface charge density and volume charge density. Surface Charge density signifies charge distribution over an area whereas linear charge density signifies charge density over a length. Linear word stands for length. Similarly, charge spread over a volume is called volume charge density. These types of charge density arise because of charge present over different dimensions. Linear charge density has charge in \[1\] dimension. Whereas surface holds in \[2\] dimension and volume in \[3\] dimensional region.