Question

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

$\bf{Options\;:}$Options:

(a) $\sf{E_A\perp E_B}E \ [tex]\less \ br /\ \greater \ A$
​
⊥E
B
​

(b) no work is done in moving a test charge $\sf{q_0}q \ [tex]\less \ br /\ \greater \ 0$
​
from B to C.
(c)$\sf{2\mid E_A \mid = 3 \mid E_B \mid}2∣E \ \textless \br/\\A$
​
∣=3∣E
B
​
∣
(d)$\sf{E_B=-E_C}E \ \[tex]texless \ br /\ \greater \ B$
​
=−E
C
​

$\boxed{\red{\bf{Note\;:}}}$
Note:
​

Among the given four alternatives out of which ANY NUMBER OF ALTERNATIVES (1 , 2 , 3 , 4) MAY BE CORRECT.

____________________________

$\sf{Explanation\:required\:} \purple{\bigstar}$
Explanationrequired★​

Answer 1

AnswEr :

Given,

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

(2,−3,−1)⟶2

(2,−4,1)⟶

$If \sf E_AE$

A

is perpendicular to $\sf E_BE$

B

, their dot product has to be zero.

A general depiction of dot product is :

$\sf a.b = a_1b_1 + a_2b_2 + a_3b_3a.b=a$

Electric Field due to a point charge :

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

E=

r

2

Kq

Now,

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

E

A

Similarly,

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

E

B

Consider only the vector parts of E_A & E_B

$\begin{gathered} \sf E_A. E_B = (2)(-1) + (-3)(-2) + (-1)(4) \\ \\ \longrightarrow \sf E_A. E_B = 6 - 6 \\ \\ \longrightarrow \sf E_A. E_B = 0 \longrightarrow (a)\end{gathered}$

=(2)(−1)+(−3)(−2)+(−1)(4)

⟶E

We notice that the respective components of $\sf E_BE$

B

and $\sf E_CE$

C

aren't equal.

Thus, option (d) is incorrect.

To bring a test charge from B to C,

$\sf W_{BC} = q_o(V_C - V_B)W </p><p>BC$

)

Also,

$\sf V = -\vec{E}.{r}V=− </p><p>E$

.r

On dotting both the vectors, we obtain :

$\sf V_B = V_C = - \dfrac{Kq}{\sqrt{21}}V </p><p>B$

Therefore,

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

W

Option (a) & (b) are correct.

Here,

Magnitude of \sf E_A = \dfrac{Kq}{14}E

and $\sf E_B = \dfrac{Kq}{21}E </p><p>B$

Dividing above expressions, we get :

$2|E_A| = 3|E_B|$

Option (c) is correct too.

Answer 2

Answer:

AnswEr :

Given,

\begin{gathered}\begin{gathered} \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} \end{gathered}\end{gathered}

r

A

(2,−3,−1)⟶2

i

^

−3

j

^

−

k

^

r

B

(−1,−2,4)⟶−

i

^

−2

j

^

+4

k

^

r

C

(2,−4,1)⟶2

i

^

−4

j

^

+

k

^

(2,−3,−1)⟶2

(2,−4,1)⟶

If \sf E_AEIfE

A

E

A

is perpendicular to \sf E_BEE

B

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_3a.b=aa.b=a

1

b

1

+a

2

b

2

+a

3

b

3

a.b=a

Electric Field due to a point charge :

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

E=

r

2

Kq

r

^

⟹

E=

∣r

2

∣

Kq

×

∣r∣

r

E=

r

2

Kq

Now,

\begin{gathered}\begin{gathered}\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}}\\ \end{gathered}\end{gathered}

E

A

=

2

2

+(−3)

2

+(−1)

2

2

Kq

×

2

2

+(−3)

2

+(−1)

2

2

i

^

−3

j

^

−

k

^

⟶E

A

=

14

14

Kq(2

i

^

−3

j

^

−

k

^

)

E

A

Similarly,

\begin{gathered}\begin{gathered}\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}}\\ \end{gathered}\end{gathered}

E

B

=

21

21

Kq(−

i

^

−2

j

^

+4

k

^

)

E

C

=

21

21

Kq(2

i

^

−4

j

^

+

k

^

)

E

B

Consider only the vector parts of E_A & E_B

\begin{gathered}\begin{gathered} \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)\end{gathered}\end{gathered}

E

A

.E

B

=(2)(−1)+(−3)(−2)+(−1)(4)

⟶E

A

.E

B

=6−6

⟶E

A

.E

B

=0⟶(a)

=(2)(−1)+(−3)(−2)+(−1)(4)

⟶E

We notice that the respective components of \sf E_BEE

B

E

B

and \sf E_CEE

C

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)W < /p > < p > BCW

BC

=q

o

(V

C

−V

B

)W</p><p>BC

)

Also,

\sf V = -\vec{E}.{r}V=− < /p > < p > EV=−

E

.rV=−</p><p>E

.r

On dotting both the vectors, we obtain :

\sf V_B = V_C = - \dfrac{Kq}{\sqrt{21}}V < /p > < p > BV

B

=V

C

=−

21

Kq

V</p><p>B

Therefore,

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

W

BC

=

21

Kqq

o

{−1−(−1)}

⟹W

BC

=0 J⟶(b)

W

Option (a) & (b) are correct.

Here,

Magnitude of \sf E_A = \dfrac{Kq}{14}E

and \sf E_B = \dfrac{Kq}{21}E < /p > < p > BE

B

=

21

Kq

E</p><p>B

Dividing above expressions, we get :

2|E_A| = 3|E_B|2∣E

A

∣=3∣E

B

∣

Option (c) is correct too.