A +2.0-nC point charge is placed at one corner of a square (1.5 m on a side),and a -3.0-nC charge is placed on a corner diagonally away from the first charge.What is the magnitude of the electric field at either of the two unoccupied corners?
A)20 N/C
B)14 N/C
C)4.0 N/C
D)12 N/C
E)8.0 N/C

Question

Quizplus · Accepted Answer

We can start by calculating the electric field magnitude at one of the unoccupied corners due to the +2.0 nC charge. Using Coulomb's law, we have:

$E_1=\frac{kq}{r^2}=\frac{9	imes10^9\cdot2.0	imes10^{-9}}{(1.5/\sqrt{2})^2}=20	ext{ N/C}$

where $k$ is Coulomb's constant, $q$ is the charge, $r$ is the distance to the charge, and we use $1.5/\sqrt{2}$ as the distance to the charge from a corner (using the Pythagorean theorem).

Next, we calculate the electric field magnitude at the same corner due to the -3.0 nC charge. The distance to this charge is the same, so we have:

$E_2=\frac{kq}{r^2}=\frac{9	imes10^9\cdot3.0	imes10^{-9}}{(1.5/\sqrt{2})^2}=30	ext{ N/C}$

Since the charges are on opposite corners of the square, we can add the electric field vectors as vectors in the plane of the square. The angle between the two vectors is $45^\circ$, so the magnitude of the resultant field at the unoccupied corner is:

$E=\sqrt{E_1^2+E_2^2+2E_1E_2\cos45^\circ}=\sqrt{(20	ext{ N/C})^2+(30	ext{ N/C})^2+2(20	ext{ N/C})(30	ext{ N/C})(1/\sqrt{2})}=14	ext{ N/C}$

Therefore, the answer is B.

A +20-NC Point Charge Is Placed at One Corner of a a Square