A point charges Q is placed at origin O. Let ${\vec{E}}_{A}, {\vec{E}}_{B} and {\vec{E}}_{C}$ represent electric fields at A, B and C respectively. If coordinates of A, B and C are respectively (1, 2, 3) m, (1, 1, -1)m and (2, 2, 2) m, then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${\vec{E}}_{A} ⊥ {\vec{E}}_{B}$

${\vec{E}}_{A} | | {\vec{E}}_{B}$

$|{\vec{E}}_{B}| = 4 |{\vec{E}}_{C}|$

$|{\vec{E}}_{B}| = 8 |{\vec{E}}_{C}|$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\vec{E} = \frac{1}{4 π \in_{0}} \frac{q}{r^{3}} \vec{r}$

${\bar{E}}_{A} = \frac{1}{4 π \in_{0}} \frac{Q}{(\sqrt{1^{2} + 2^{2} + 3^{2}})} (\hat{i} + 2 \hat{j} + 3 \hat{k})$

$\begin{array}{l} {\bar{E}}_{A} = \frac{1}{4 π \in_{0}} . \frac{Q}{14 \sqrt{14}} (\hat{i} + 2 \hat{j} + 3 \hat{k}) \\ {\vec{E}}_{B} = \frac{1}{4 π \in_{0}} \frac{Q}{{\sqrt{(1^{2} + 1^{2} + 1^{2})}}^{3}} (\hat{i} + \hat{j} - \hat{k}) = \frac{1}{4 π \in_{0}} \frac{Q}{3 \sqrt{3}} (\hat{i} + \hat{j} - \hat{k}) \\ {\vec{E}}_{C} = \frac{1}{4 π \in_{0}} \frac{Q}{{(\sqrt{2^{2} + 2^{2} + 2^{2}})}^{3}} (2 \hat{i} + 2 \hat{j} + 2 \hat{k}) = \frac{1}{4 π \in_{0}} \frac{Q}{12 \sqrt{12}} (2 \hat{i} + 2 \hat{j} + 2 \hat{k}) \end{array}$