# ElectrostaticsThree charges are placed at the corners of a 45-45-90 degree triangle.  Let q45_1 = +3.2*10^-19 C, q45_2 = +3.2*10^-19 C and q90 = +6.4*10^-19 C.  If the legs of the triangle are of...

Electrostatics

Three charges are placed at the corners of a 45-45-90 degree triangle.  Let q45_1 = +3.2*10^-19 C, q45_2 = +3.2*10^-19 C and q90 = +6.4*10^-19 C.  If the legs of the triangle are of length 1 meter, find the magnitude of the electric field at point P, located at the center of the hypothenus .

txmedteach | High School Teacher | (Level 3) Associate Educator

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The electric field at a point is the sum of all coulombic forces on a 1 C charge at that point. The electric field from one point charge is found with the following equation:

E = Q/(4*pi*epsi_0*r^2)

Here, Q is the known charge, epsi_0 is the dielectric constant of a vacuum, and r is the distance between the charge and the point at which we're calculating the electric field.

This quantity is a vector, too, where the direction is toward the charge if the it is negative and away from the charge if it is positive.

So, all we're doing is calculating this quantity for 3 charges.

Actually, let's stop right there, because we can make this much easier on ourselves. The two charges at the 45 degree angles of the triangle are of equal magnitude. Therefore, their action on the center of the hypotenuse cancels out because the electric fields generated by these two are equal and opposite. Therefore, the only influence on the electric field at point P is the charge at the right angle.

In order to determine the electric field at the point, we need to first determine the distance between P and the right angle of the triangle. Thankfully, this is a 45-45-90 triangle, and the length of the segment from the right angle and middle of the hypotenuse is 1/2 the length of the hypotenuse. Because the legs are each 1 meter, we can say the hypotenuse is sqrt(2) meters. Therefore, the "r" that we will use in the electric field equation is sqrt(2)/2.

So, we know the following information for our equation:

Q = 6.4*10^-19 C

r = sqrt(2)/2 m

epsi_0 = 8.85*10^-12 C^2/Nm^2 (This number is a constant)

Let's calculate the electric field magnitude:

E = Q/(4*pi*epsi_0*r^2)

= 6.4*10^-19/(4*3.14*8.85*10^-12*1/4)

E = 2.30*10^-8 N/C

Again, this is our final magnitude because the field from the other charges cancels out.

Notice that our units are in terms of Newtons per Coulomb. This unit is to be expected because to formula to find the force on a charge by an electric field is the following:

F = Eq

where q is the charge of the particle on which we're finding the force.

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