(a) Use Gauss's law to derive the expression for the electric field $\vec{E}$ due to a straight uniformly charged infinite line of charge density $\lambda\ \text{C m}^{-1}$. (b) Draw a graph to show the variation of $E$ with perpendicular distance $r$ from the line of charge. (c) Find the work done in bringing a charge $q$ from perpendicular distance $r_1$ to $r_2$ ($r_2 > r_1$).

(a) $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$. (b) $E \propto \dfrac{1}{r}$ (a rectangular hyperbola). (c) $W = \dfrac{q\lambda}{2\pi\varepsilon_0}\ln\!\left(\dfrac{r_2}{r_1}\right)$.

CBSE Class 12 · 2018 · ElectrostaticsLong Answer

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(a) Use Gauss's law to derive the expression for the electric field

E

due to a straight uniformly charged infinite line of charge density

λ C m^{- 1}

.
(b) Draw a graph to show the variation of

E

with perpendicular distance

r

from the line of charge.
(c) Find the work done in bringing a charge

q

from perpendicular distance

r_{1}

r_{2}

(

r_{2} > r_{1}

Solution

(a) With a coaxial Gaussian cylinder (radius

r

, length

ℓ

E (2 π r ℓ) = \frac{λ ℓ}{ε _{0}} \Rightarrow E = \frac{λ}{2 π ε _{0} r}

(b) Since

E \propto \frac{1}{r}

, the graph of

E

against

r

is a rectangular hyperbola.
(c) Work done from

r_{1}

r_{2}

W = q \int_{r_{1}}^{r_{2}} E d r = \frac{q λ}{2 π ε _{0}} \int_{r_{1}}^{r_{2}} \frac{d r}{r}

W = \frac{q λ}{2 π ε _{0}} ln (\frac{r _{2}}{r _{1}})

CBSE Class 12 Board · 2018 · Subjective

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