You know that a point charge is located somewhere along the x-axis. when you measure the electric field at 2.00 m the result is 3.000 n/c and points left while the electric field at 5.00 m is 0.750 n/c also pointing left. (a) where is the charge located?
(b) what is the size of the charge, including sign?

3 nC.

Explanation:

Let the charge be Q and it is situated at distance x from first point and

x + 3 from second point.

Applying formula for electric field we have , for first point

k Q / x² = 3

Applying formula for electric field we have , for second  point

k Q / (x +3)² = .750

Dividing ,

(x +3)² / x² = 3/.750

= 4

(x +3) / x = 2

x = 3

Therefore the charge must be present at - ve 1 m . That is why its distance from x =2 m on + ve x axis is 3 m

Size of charge

k Q / x² = 3

Q = 3 x² / k

= (3 x 3²) / 9 x 10⁹

=   3 x 10⁻⁹

= 3 nC.

your answer is c! hope this : )

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y