Question

A charge q is situated at the origin. Let \sf{E_A}E
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) $\sf{E_A\perp E_B}E$
A
​
⊥E
B
​

(b) no work is done in moving a test charge \sf{q_0}q
0
​
from B to C.
(c)$\sf{2\mid E_A \mid = 3 \mid E_B \mid}2∣E$
A
​
∣=3∣E
B
​
∣
(d)$\sf{E_B=-E_C}E$
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.​

Answer 1

Answer:

-E)) tex is equal to 3B

Step-by-step explanation:

I expert my answar is right

Answer 2

ANSWER⤵

-E)) tex is =3B

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