Question

The electric field at (30,30) cm due to a charge of -8nC at the origin in N/C is
(A) $- 400(\widehat i + \widehat j)$
(B) $400(\widehat i + \widehat j)$
(C) $- 200\sqrt 2 (\widehat i + \widehat j)$
(D) $200\sqrt 2 (\widehat i + \widehat j)$

Answer 1

The electric field caused by a stationary charge in free space is defined as the force per unit charge i.e. the force experienced by a test charge when it enters the field.

Formula used:

$$\overrightarrow F = \dfrac{1}{{4\pi {\\_o}}}\dfrac{{Q{q_o}}}{{{r^3}}}\overrightarrow r \\\ \overrightarrow E = \dfrac{{\overrightarrow F }}{{{q_o}}} \\\ \overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{Q}{{{r^3}}}\overrightarrow r \\\$$

Where,
${q_o}$is the test charge
Q is the charge which caused the electric field at the first place
$\overrightarrow r$is the displacement vector from the charge Q to the point where electric field has to be calculated

Complete step by step answer:
When a charged particle is placed in free space it produces an electric field in its vicinity which generally tends to zero at infinity. This happens with every particle whether charged or not.
Gravitational particle gravitons produce a gravitational field, photons produce electromagnetic fields, phonons also have their own field which creates vibrations(disturbances) in the medium and so on.
So, in vector form the electric field is given as:
$\overrightarrow E = \dfrac{1}{{4\pi {\\_o}}}\dfrac{Q}{{{r^3}}}\overrightarrow r$
According to the question,

$$Q = - 8nC \\\ r = 0.3\overrightarrow i + 0.3\overrightarrow j \\\ \left| {\overrightarrow r } \right| = \sqrt {{{0.3}^2} + {{0.3}^2}} = 0.3\sqrt 2 m \\\$$

The units have been converted accordingly here (cm to m). Substituting the above values in the electric field expression, we have,

$$\overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{( - 8 \times {{10}^{ - 9}})}}{{{{0.3}^3} \times 2\sqrt 2 }}0.3(\overrightarrow i + \overrightarrow j ) \\\ \Rightarrow - \dfrac{{9 \times {{10}^9} \times 8 \times {{10}^{ - 9}}}}{{{{0.3}^2} \times 2\sqrt 2 }}(\overrightarrow i + \overrightarrow j ) \\\ \overrightarrow E = - 200\sqrt 2 (\overrightarrow i + \overrightarrow j )N{C^{ - 1}} \\\$$

Therefore, the electric field at (30,30) cm due to a charge of -8nC at the origin is $\overrightarrow E = - 200\sqrt 2 (\overrightarrow i + \overrightarrow j )N{C^{ - 1}}$
The correct answer is option C.

Note: Sometimes in a hurry, students do not convert units in the first place and get wrong answers. This leads to rechecking the answer and a lot of time is wasted therefore read the question properly. Also, you may get the vector formula for electric field or force wrong. What happens is that instead of $\overrightarrow r$ you may write $\widehat r$which is correct but you forget to change ${r^3}$to ${r^2}$as $\overrightarrow r = \left| {\overrightarrow r } \right|\widehat r$.