Two equal positive charges q are placed at the opposite vertices of a square of side a. Determine and show on a diagram the electric field created by these three charges at the fourth vertex of the square.

Question

Two equal positive charges q are placed at the opposite vertices of a square of side a. Determine and show on a diagram the electric field created by these three charges at the fourth vertex of the square.

Thanks.

Answer 1

Two charges on three vertices?

anyway
do
E = k q/r^2 in the direction from each charge to the spot and add the vectors.

Answer 2

Sorry I missed a sentence

A charge -q is placed at the third vertex

Answer 3

To determine the electric field at the fourth vertex of the square created by the two equal positive charges at the opposite vertices, we can use the principle of superposition.

Let's label the charges as Q1 and Q2, and the distance between the charges as R.

First, we calculate the electric field due to Q1 at the fourth vertex. The electric field due to a point charge can be calculated using Coulomb's law:

Electric field due to Q1 = k * Q1 / R^2

where k is the Coulomb's constant (8.99 × 10^9 N·m^2/C^2).

Next, we calculate the electric field due to Q2 at the fourth vertex. The electric field due to a point charge is given by:

Electric field due to Q2 = k * Q2 / R^2

Since the charges are equal (q = Q1 = Q2) and the distances are equal from the fourth vertex to the charges (a), we can simplify the calculations:

Electric field due to Q1 = Electric field due to Q2 = k * q / a^2

Now, to find the total electric field at the fourth vertex, we use the principle of superposition. In this case, since the electric fields due to Q1 and Q2 have the same magnitude and direction, we can add them up:

Total electric field = Electric field due to Q1 + Electric field due to Q2 = 2 * (k * q / a^2)

To show this visually on a diagram, consider a square with the charges Q1 and Q2 at the two opposite vertices. At the fourth vertex, draw two arrows representing the electric fields due to Q1 and Q2, both pointing towards the fourth vertex. The length and direction of the arrows should represent the magnitude and direction of the calculated electric field.

Note that the direction of the electric field can be positive (towards the charge) or negative (away from the charge), depending on the sign of the charges. In this case, since both charges are positive, the electric field at the fourth vertex will be positive, directed towards the charges.

I hope this explanation helps! Let me know if you have any more questions.