If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?
A) 33 N/C
B) 31 N/C
C) 24 N/C
D) 19 N/C
E) 13 N/C

Question

If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P? 
A) 33 N/C 
B) 31 N/C 
C) 24 N/C 
D) 19 N/C 
E) 13 N/C

Quizplus · Accepted Answer

The magnitude of the electric field at point P can be calculated using the formula:

$E = \frac{kQ}{r^2}$

where k is the Coulomb constant, Q is the charge, and r is the distance from the charge to the point P.

In this case, we have:

$k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$
$Q = 16 	ext{ nC} = 16 	imes 10^{-9} 	ext{ C}$
$r = \sqrt{(3.0 	ext{ m})^2 + (4.0 	ext{ m})^2} = 5.0 	ext{ m}$

Plugging these values into the equation, we get:

$E = \frac{(9 	imes 10^9 	ext{ N m}^2/	ext{C}^2)(16 	imes 10^{-9} 	ext{ C})}{(5.0 	ext{ m})^2} = 23.04 	ext{ N/C}$

Rounding to two significant figures, we get $E = 24 	ext{ N/C}$, which is answer choice C.

If Q = 16 NC, a = 3