Question

Hola brainlieans !Here is a challenging question for you

1) Find the electric field at a point located at 2cm charge of 1nC.​

Answer 1

Given :

Magnitude of charge = 1 nC

To Find :

Electric field at a point located at 2 cm from the point charge.

Solution :

★ Electric field is the region around charged particle or charged body in which if another charge is placed, it experiences electrostatic force.

• It is a vector quantity. Its direction is the same as the force experienced by positive charge.
• Direction of electric field due to positive charge is always away from it while due to negative charge always towards it.

Electric field intensity at distance of r from point charge q is given by

$:\implies\:\underline{\boxed{\bf{\purple{E=\dfrac{kq}{r^2}}}}}$

where k = 9 × 10⁹ N m² / C²

By substituting the given values;

$\sf:\implies\:E=\dfrac{(9\times 10^9)\cdot{(1\times10^{-9})}}{(2\times 10^{-2})^2}$

$\sf:\implies\:E=\dfrac{9}{4\times10^{-4}}$

$:\implies\:\underline{\boxed{\bf{\orange{E=2.25\times 10^4\:NC^{-1}}}}}$

Answer 2

Solution:

Using the formula,

$\bf \leadsto \: E = \frac{ \lambda}{2\pi \epsilon or}$

$\bf \leadsto\epsilon \:o = 8.85 \times {10}^{ - 12} \: f. {m}^{ - 1} (farad \: perimeter)$

$\leadsto \bf n = 2cm = 0.02m$

$\leadsto \bf \: \pi = 3.14$

$\bf\lambda = {10}^{ - 7} {cm}^{ - 1} (coulomb \: per \: meter)$

$\bf \: Now, \: E = \frac{ {10}^{ - 7} }{2 \times 3.14 \times 8.85 \times {10}^{ - 12} \times 0.02 }$

$\bf \leadsto \frac{1}{1.11156} \times \frac{ {10}^{2 - 7} }{ {10}^{ - 12} }$

$\bf \leadsto \frac{1}{( \frac{118}{100}) } \times \frac{ {10}^{5} }{1} = \frac{100}{8} \times \frac{ {10}^{5} }{1} = \frac{ {10}^{7} }{118}$

$\bf \leadsto 118 \times {10}^{ - 7}$

$\bf \leadsto1.18 \times {10}^{ - 9} \: \: \: \: \: \: \lambda \: . {C}^{-1}$