Question
A rod of length $$2.4\,m$$ and radius $$4.6\,mm$$ carries a negative charge of $$ - 4.2 \times {10^{ - 7}}C$$ spread uniformly over it surface. The electric field near the mid-point of the rod, at a point on its surface is
A.
$$ - 8.6 \times {10^5}N{C^{ - 1}}$$
B.
$$8.6 \times {10^4}N{C^{ - 1}}$$
C.
$$ - 6.7 \times {10^5}N{C^{ - 1}}$$
D.
$$6.7 \times {10^4}N{C^{ - 1}}$$
Answer :
$$ - 6.7 \times {10^5}N{C^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Here,}}\,\ell = 2.4\,m,r = 4.6\,mm = 4.6 \times {10^{ - 3}}\,m \cr
& q = - 4.2 \times {10^{ - 7}}C \cr} $$
Linear charge density, $$\lambda = \frac{q}{\ell }$$
$$ = \frac{{ - 4.2 \times {{10}^{ - 7}}}}{{2.4}} = - 1.75 \times {10^{ - 7}}C{m^{ - 1}}$$
Electric field, $$E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}$$
$$\eqalign{
& = \frac{{ - 1.75 \times {{10}^{ - 7}}}}{{2 \times 3.14 \times 8.854 \times {{10}^{ - 12}} \times 4.6 \times {{10}^{ - 3}}}} \cr
& = - 6.7 \times {10^5}\,N{C^{ - 1}} \cr} $$