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

We can use the equation for the magnitude of the electric field due to a point charge:

$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$

where $\epsilon_0$ is the permittivity of free space, $Q$ is the charge, and $r$ is the distance from the charge.

In this case, we want to find the magnitude of the electric field at point P, which is a distance $d$ away from the charge Q:

$d = \sqrt{a^2 + b^2} = \sqrt{(3.0 	ext{ m})^2 + (4.0 	ext{ m})^2} = 5.0 	ext{ m}$

Plugging in the values, we get:

$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2} = \frac{1}{4\pi(8.85	imes10^{-12} 	ext{ N}^{-1}	ext{m}^{-2})}\frac{16	imes10^{-9} 	ext{ C}}{(5.0 	ext{ m})^2}$

Solving for $E$, we get:

$E = 23.53 	ext{ N/C}$

Rounding to the nearest whole number, we get:

$E \approx 24 	ext{ N/C}$

Therefore, the best choice is C.

If Q = 16 NC, a = 3