Question

A charge q is situated at the origin. Let $\sf{E_A}$, $\sf{E_B}$ and $\sf{E_C}$ be the electric fields at the points A(2, -3, -1), B(-1, -2, 4) and 2, -4, 1). Therefore,

$\bf{Options\;:}$

(a) $\sf{E_A\perp E_B}$
(b) no work is done in moving a test charge $\sf{q_0}$ from B to C.
(c)$\sf{2\mid E_A \mid = 3 \mid E_B \mid}$
(d) $\sf{E_B=-E_C}$

$\boxed{\red{\bf{Note\;:}}}$
Among the given four alternatives out of which ANY NUMBER OF ALTERNATIVES (1 , 2 , 3 , 4) MAY BE CORRECT.

____________________________

$\sf{Explanation\:required\:} \purple{\bigstar}$

Answer 1

AnswEr :

Given,

$\sf r_A (2,-3,-1) \longrightarrow 2 \hat{i} - 3 \hat{j} - \hat{k} \\ \sf r_B (-1,-2,4) \longrightarrow - \hat{i} - 2 \hat{j} +4 \hat{k} \\ \sf r_C (2,-4,1) \longrightarrow 2 \hat{i} - 4 \hat{j} + \hat{k}$

If $\sf E_A$ is perpendicular to $\sf E_B$, their dot product has to be zero.

A general depiction of dot product is :

$\sf a.b = a_1b_1 + a_2b_2 + a_3b_3$

Electric Field due to a point charge :

$\sf \: E = \dfrac{Kq}{r {}^{2} } \: \hat{r} \\ \\ \implies \boxed{ \boxed{ \sf \: E = \dfrac{Kq}{ \mid r {}^{2} \mid} \: \times \dfrac{ \vec{r}}{ |r| } }}$

Now,

$\sf E_A = \dfrac{Kq}{{\sqrt{2^2 + (-3)^2 + (-1)^2}}^{2}} \times \dfrac{2\hat{i} - 3 \hat{j} - \hat{k}}{\sqrt{2^2 + (-3)^2 + (-1)^2}} \\ \\ \longrightarrow \sf E_A = \dfrac{Kq(2\hat{i} - 3 \hat{j} - \hat{k})}{14 \sqrt{14}}$

Similarly,

$\sf E_B = \dfrac{Kq(- \hat{i} - 2 \hat{j} +4 \hat{k})}{21 \sqrt{21}} \\ \\ \sf E_C = \dfrac{Kq(2\hat{i} -4\hat{j} + \hat{k})}{21 \sqrt{21}}$

Consider only the vector parts of E_A & E_B

$\sf E_A. E_B = (2)(-1) + (-3)(-2) + (-1)(4) \\ \\ \longrightarrow \sf E_A. E_B = 6 - 6 \\ \\ \longrightarrow \sf E_A. E_B = 0 \longrightarrow (a)$

We notice that the respective components of $\sf E_B$ and $\sf E_C$ aren't equal.

Thus, option (d) is incorrect.

To bring a test charge from B to C,

$\sf W_{BC} = q_o(V_C - V_B)$

Also,

$\sf V = -\vec{E}.{r}$

On dotting both the vectors, we obtain :

$\sf V_B = V_C = - \dfrac{Kq}{\sqrt{21}}$

Therefore,

$\sf W_{BC} = \dfrac{Kqq_o}{\sqrt{21}}\{- 1 - (-1) \} \\ \\ \implies \sf W_{BC} = 0 \ J \longrightarrow (b)$

Option (a) & (b) are correct.

Here,

Magnitude of $\sf E_A = \dfrac{Kq}{14}$ and $\sf E_B = \dfrac{Kq}{21}$

Dividing above expressions, we get :

2|E_A| = 3|E_B|

Option (c) is correct too.

Answer 2

The electric flux through the whole cube:

ϕ= ε 0Q

Flux through each face will be one sixth of the total:

ϕ= 6ε 0Q