Question

Question No S
A negative point charge (10 to the
power minus 8 in coulomb is situated
in air at the origin of a rectangular co-
ordinate system. What is the electric
field strength at a point 3 meter away
on positive x-axis?
Answer
A. 10 N/C and directed to positive
x-axis
B. 10 N/C and directed to negative
xas
C.
30 N/C and directed to positive​

Answer 1

Given:

A negative point charge 10^(-8) Coulomb is situated in air at the origin of a rectangular co-ordinate system.

To find:

Electrostatic Field Intensity at a point 3 metres on positive X axis?

Calculation:

Let field intensity be E :

$\rm \therefore \: \vec{E} = |E| \hat{i}$

$\rm \implies \: \vec{E} = \dfrac{kq}{ {x}^{2} } \: \hat{i}$

$\rm \implies \: \vec{E} = \dfrac{q}{ 4\pi \epsilon_{0}{x}^{2} } \: \hat{i}$

$\rm \implies \: \vec{E} = \dfrac{ (- {10}^{ - 8}) \times (9 \times {10}^{9} )}{{(3)}^{2} } \: \hat{i}$

$\rm \implies \: \vec{E} = \dfrac{ 90}{9 } \:( - \hat{i})$

$\rm \implies \: \vec{E} = 10 \: N/C\:( - \hat{i})$

• $(-\hat{i})$ means that field intensity factor is directed along the negative X axis.

So, field intensity will be 10 N/C directed towards -X axis (OPTION B).