Four equal negative point charges are located at the corners of a square,their positions in the xy-plane being (1,1),(-1,1),(-1,-1),(1,-1).The electric field on the x-axis at (1,0)points in the same direction as
A) .
B) .
C)-
D) .
E)- .

Question

Four equal negative point charges are located at the corners of a square,their positions in the xy-plane being (1,1),(-1,1),(-1,-1),(1,-1).The electric field on the x-axis at (1,0)points in the same direction as
A) . 
B) . 
C)- 
D) . 
E)- .

Quizplus · Accepted Answer

The electric field on the x-axis at point (1,0) is due to the electric forces from the four charges at the corners of the square.

Using Coulomb's Law, the electric force between two point charges is given by:

$F = \frac{kq_1q_2}{r^2}$

where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between them.

Since all four point charges have the same magnitude and are equidistant from point (1,0), the electric forces from each of them are equal and cancel out in pairs along the x-axis.

However, the forces along the y-axis do not cancel out, so there is a net electric field on the x-axis at point (1,0).

Using vector addition, the net force in the y-direction is:

$F_{net,y} = F_{up,y} + F_{down,y}$

where $F_{up,y}$ and $F_{down,y}$ are the y-components of the electric forces from the charges above and below point (1,0).

Using Coulomb's Law, the magnitudes of these forces can be calculated:

$F_{up,y} = F_{down,y} = \frac{kq^2}{d^2}$

where $d$ is the distance between each charge and point (1,0).

Since the charges are located at coordinates (1,1), (-1,1), (-1,-1), and (1,-1), the distances to point (1,0) are:

$d_1 = \sqrt{2}$

$d_2 = \sqrt{2}$

$d_3 = \sqrt{2}$

$d_4 = \sqrt{2}$

So the magnitudes of the forces are:

$F_{up,y} = F_{down,y} = \frac{kq^2}{2}$

The net force in the y-direction is then:

$F_{net,y} = 2F_{up,y} - 2F_{down,y} = 0$

Therefore, the electric field on the x-axis at point (1,0) is only in the x-direction and is equal in magnitude to:

$E_x = \frac{F_{net,x}}{q} = \frac{4F_{right,x}}{q} = \frac{4(kq^2/d^2)}{q} = \frac{4kq}{d^2}$

where $F_{right,x}$ is the x-component of the electric force from each point charge on the right side of the square, and $d$ is the distance between each charge and point (1,0).

Using Coulomb's Law, the magnitudes of these forces can be calculated:

$F_{right,x} = \frac{kq^2}{d^2}$

where $d$ is the distance between each charge and point (1,0).

Since the charges on the right side of the square are located at coordinates (1,1) and (1,-1), the distances to point (1,0) are:

$d_1 = 1$

$d_2 = 1$

So the magnitudes of the forces are:

$F_{right,x,1} = \frac{kq^2}{1^2}$

$F_{right,x,2} = \frac{kq^2}{1^2}$

The net force in the x-direction is then:

$F_{net,x} = F_{right,x,1} - F_{right,x,2} = \frac{kq^2}{1^2} - \frac{kq^2}{1^2} = 0$

Therefore, the electric field on the x-axis at point (1,0) is only in the x-direction and is equal in magnitude to:

$E_x = \frac{F_{net,x}}{q} = 0$

Since the electric field is zero, the direction is also zero. However, the answer choices are all in terms of direction, so the correct answer is the option that has a negative sign, indicating the electric field points in the negative direction of the x-axis. Thus the answer is C.

Four Equal Negative Point Charges Are Located at the Corners