JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 13)
Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/m lying along the z-axis in free space.
0.424 nC m$$-$$2
4.0 nC m$$-$$2
47.88 C/m
0.07 nC m$$-$$2
Explanation
Electric field due to wire is given by $$E = {{2k\lambda } \over r}$$
Electric field with surface charge density $$E = {\sigma \over {{ \in _0}}}$$
$${{2k\lambda } \over r} = {\sigma \over {{ \in _0}}}$$
$$ \Rightarrow 2{1 \over {4\pi { \in _0}}}{\lambda \over r} = {\sigma \over {{ \in _0}}}$$
$$ \Rightarrow {8 \over {2 \times 3.14 \times 3}} = \sigma $$
$$ \Rightarrow \sigma = 0.424$$ n Cm$$-$$2
Electric field with surface charge density $$E = {\sigma \over {{ \in _0}}}$$
$${{2k\lambda } \over r} = {\sigma \over {{ \in _0}}}$$
$$ \Rightarrow 2{1 \over {4\pi { \in _0}}}{\lambda \over r} = {\sigma \over {{ \in _0}}}$$
$$ \Rightarrow {8 \over {2 \times 3.14 \times 3}} = \sigma $$
$$ \Rightarrow \sigma = 0.424$$ n Cm$$-$$2
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