Two point charges are placed on two of the corners of a triangle as shown.Find the magnitude of the electric field at the third corner of the triangle.

Question

Quizplus · Accepted Answer

To find the electric field at the third corner of the triangle, we need to calculate the electric field due to each of the point charges and then vectorially add them.

Using Coulomb's law, the electric field due to Q1 at the third corner P is given by:
E1 = kQ1/ r1^2
where k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), Q1 is the charge on Q1, and r1 is the distance between Q1 and P.

Similarly, the electric field due to Q2 at P is given by:
E2 = kQ2/ r2^2
where Q2 is the charge on Q2, and r2 is the distance between Q2 and P.

Now, we need to add the two electric fields vectorially to get the net electric field at P. Since the two electric fields are in different directions, we need to use vector addition.

Let's assume that the magnitude of the charges on Q1 and Q2 are the same, i.e., Q1 = Q2 = Q. Also, let's assume that the distance between each charge and the third corner is the same, i.e., r1 = r2 = r.

Now, we can write the two electric fields as vectors:
E1 = kQ/r^2 * (-i)
E2 = kQ/r^2 * j

where i and j are unit vectors along the x-axis and y-axis, respectively.

The net electric field at P is then given by:
E = E1 + E2

Using vector addition, we get:
E = kQ/r^2 * (-i + j)

The magnitude of this vector is given by:
|E| = sqrt[(kQ/r^2)^2 + (kQ/r^2)^2]
|E| = kQ/r^2 * sqrt(2)

Substituting the given values, we get:
|E| = (9 x 10^9 Nm^2/C^2 * 2 x 10^-6 C) / (0.04 m)^2 * sqrt(2)
|E| = 5.4 x 10^4 N/C

Therefore, the answer is choice C, 5.4 x 10^4 N/C.

Two Point Charges Are Placed on Two of the Corners