Four identical positive charges q are placed at the four vertices of a square of side a. Determine the electric field:
A) at the center of the square
B) at the midpoint of one of its sides
A - ROTFL - everything is equal and opposite to everything else
B - now we have a problem
the two on the same side cancel each other but the two on the opposite charge try to push away at 90 degrees to the side
distance from each = sqrt (a^2 + .25 a^2) = a sqrt (5/4)
component perpendicular to side
= [2 k q /(5 a^2 /4)] sin theta
where sin theta = 1/[ sqrt5/4]
= 2 k q/[ a^2 (5/4)^1.5 ]
To determine the electric field at the center of the square, we can consider the electric field due to each charge individually and then sum them up vectorially.
Let's call the positive charges q1, q2, q3, and q4, and consider a charge q at the center of the square.
A) Electric field at the center of the square:
- The electric fields due to q1 and q2 will point directly away from the center, with magnitudes given by Coulomb's law: E1 = kq/r^2, where r is the distance between q1/q2 and the center of the square (which is a/2).
- The electric fields due to q3 and q4 will point directly towards the center, with magnitudes given by Coulomb's law: E3 = kq/r^2, where r is the distance between q3/q4 and the center of the square (which is a/2).
Since the charges are placed at the vertices of a square, the distances r are equal for all four charges.
Considering that the fields due to opposite charges (+q and -q) cancel each other out, we can sum the fields due to q1 and q2, and the fields due to q3 and q4 separately.
The net electric field at the center of the square, E_net, will be the sum of the two individual electric fields:
E_net = E1 + E2 + E3 + E4
Since the magnitudes of the electric fields due to q1, q2, q3, and q4 are all the same, we have:
E_net = 2E1 + 2E3
Substituting the values, we get:
E_net = 2(kq/r^2) + 2(kq/r^2) = 4(kq/r^2)
So, the electric field at the center of the square is 4 times the electric field due to each individual charge.
B) Electric field at the midpoint of one of the sides of the square:
At the midpoint of one of the sides of the square, the electric field due to opposite charges on the adjacent vertices will cancel each other out.
The only non-zero contributions to the electric field come from the charges at the remaining two vertices.
Let's call the midpoint charge q_mid.
The electric field due to a single charge q at one vertex of the square at the midpoint is given by Coulomb's law: E_mid = kq/r^2, where r is the distance between the charge and the midpoint.
Since there are two charges contributing to the electric field at the midpoint, we sum the two electric fields:
E_net_mid = E_mid1 + E_mid2
Substituting the values, we get:
E_net_mid = (kq/r^2) + (kq/r^2) = 2(kq/r^2)
So, the electric field at the midpoint of one of the sides of the square is 2 times the electric field due to each individual charge.
To determine the electric field at a point due to the charges placed at the vertices of a square, we can use the principle of superposition. This principle states that the total electric field at a point due to multiple charges is the vector sum of the electric fields created by each individual charge.
A) To find the electric field at the center of the square (let's call it point P), we need to calculate the electric field created by each of the four charges at P and then sum them up.
1. Calculation for the electric field created by one charge at P:
Since all charges are identical, we can just focus on one charge. The electric field created by a point charge q at a distance r is given by Coulomb's Law:
E = k * (q / r^2),
where k is the electrostatic constant (k ≈ 8.99 × 10^9 N m^2/C^2).
Considering symmetry, the electric field created by one charge at the center of a square (P) is directed along the diagonal of the square. The distance from one corner of the square to the center is r = a/√2.
So, the electric field created by one charge at P is:
E1 = k * (q / (a / √2)^2)
= k * (q / (a^2 / 2))
= 2 * k * (q / a^2).
2. Summing up the electric fields from all four charges:
Since the square has identical charges at each vertex and they are equidistant from the center, the total electric field at P is:
E_total = 4 * E1
= 4 * (2 * k * (q / a^2))
= 8 * k * (q / a^2).
Therefore, the electric field at the center of the square is 8 times the electric field created by a single charge, in the direction along the diagonal of the square.
B) To find the electric field at the midpoint of one of the sides of the square, let's select one of the sides (let's call it AB) and its midpoint (M). We need to determine the electric field created by each of the four charges at M and then sum them up.
1. Calculation for the electric field created by one charge at M:
The distance between the midpoint of a side and one of the charges is r = a/2.
So, the electric field created by one charge at M is:
E1 = k * (q / (a/2)^2)
= 4 * k * (q / a^2).
2. Summing up the electric fields from all four charges:
Again, since the charges are identical and equidistant from the midpoint, the total electric field at M is:
E_total = 4 * E1
= 4 * (4 * k * (q / a^2))
= 16 * k * (q / a^2).
Therefore, the electric field at the midpoint of one of the sides is 16 times the electric field created by a single charge, directed perpendicular to the side of the square.
Note: The direction of the electric field is important, so be sure to include it in your answer.