4 charges Q1=2C, Q2=3C, Q3=3C & Q4=-4C are placed at the vertices of a square of side 10cm. find the force on 5th charge Q5=1C placed at the center of square.

13.5×10

To find the force on the 5th charge (Q5) placed at the center of the square, we can apply Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Given:
Charge Q1 = 2C,
Charge Q2 = 3C,
Charge Q3 = 3C,
Charge Q4 = -4C,
Side of the square = 10cm.

Step 1: Calculate the distance between the center of the square (where charge Q5 is placed) and each of the charges at the vertices of the square. Since the square has side length 10cm, the distance between the center and any vertex is half of the side length, which is 10cm / 2 = 5cm.

Step 2: Calculate the force between Q5 and each of the charges Q1, Q2, Q3, and Q4 using Coulomb's Law. Coulomb's Law equation is:

F = (k * |Q1 * Q5|) / d^2

Where F is the force, k is the electrostatic constant (9 × 10^9 N⋅m^2/C^2), Q1 and Q5 are the charges, and d is the distance between them.

For Q1:
F1 = (k * |2C * 1C|) / (5cm)^2

For Q2:
F2 = (k * |3C * 1C|) / (5cm)^2

For Q3:
F3 = (k * |3C * 1C|) / (5cm)^2

For Q4:
F4 = (k * |-4C * 1C|) / (5cm)^2

Note: The charge Q4 is negative, so the force will be attractive instead of repulsive.

Step 3: Calculate the total force exerted on charge Q5 by adding the individual forces:

Total Force on Q5 = F1 + F2 + F3 + F4

Step 4: Calculate the direction of the force. Since the charges Q1, Q2, and Q3 have the same magnitude (3C), their forces will have equal magnitudes but act along different directions due to the symmetry of the square. The forces F1, F2, and F3 will form an equilibrium, canceling each other out. The force F4 due to Q4 will be the net force acting on Q5, pulling it towards Q4.

Hence, the force on the 5th charge Q5 placed at the center of the square will be given by the formula and calculations described above.