Question: Q. 2. (i) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λC/m.

(ii) Draw a graph to show the variation of E with perpendicular distance r from line of charge.

(iii) Find the work done in bringing a charge q from perpendicular distance r1 to r2(r2>r1).

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Solution:

Ans. (i) Derivation of the expression for electric field E

(ii) Graph to show the required variation of the electric field

(iii) Calculation of work done 1

(i)

To calculate the electric field, imagine a cylindrical Gaussian surface, since the field is everywhere radial, flux through two ends of the cylindrical Gaussian surface is zero.

1/2

At cylindrical part of the surface electric field E is normal to the surface at every point and its magnitude is constant.

Therefore flux through the Gaussian surface = Flux through the curved cylindrical part of the surface,

(i)=E×2πrl

Applying Gauss’s Law

Flux,

ϕ=qenclosed ε0

Total charge enclosed = Linear charge density ×l

(ii)=λl ϕ=λlε0

Using Equations (i) & (ii)

E×2πrl=λlε0 E=λ2πε0r

In vector notation, E=λ2πε0rn^

(where n^ is a unit vector normal to the line charge) (ii) The required graph is as shown :

(iii) Work done in moving the charge ’ q ’ through a small displacement ’ dr '

dW=Fdr dW=qEdr =qEdrcos0 dW=q×λ2πε0rdr

Work done in moving the given charge from r1 to r2 (r2>r1)

$$ \begin{aligned} & W=\int_{r_{1}}^{r_{2}} d W=\int_{r_{1}}^{r_{2}} \frac{\lambda g d r}{2 \pi \varepsilon_{0} r} \ & W=\frac{\lambda q}{2 \pi \varepsilon_{0}}\left[\log {e} r{2}-\log {e} r{1}\right] \quad 1 / 2 \ & W=\frac{\lambda q}{2 \pi \varepsilon_{0}}\left[\log {e} \frac{r{2}}{r_{1}}\right] \end{aligned} $$

[CBSE Marking Scheme, 2018]



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