Problem in physic of electric charge

[lovelycecila]

consider a circular clock face of radius= 50cm.a charge of 3nC is placed at the 3 o'clockpoint (a distane of 50cm from the center of the face). a charge of 6nC is placed at the 6 o'clockpoint.9nC is placed at the 9 o'clockpoint and 12nC is placed at the 12 o'clockpoint.
what is the electric field in the centre of the clock

 

[Defennder]

Well, the net electric field is due to the vector superposition of all the electric charges. So how would you use that to do this question?

 

[thomate1]

The problem described in this scenario involves the principle of electric charge and its effects on the surrounding space. Specifically, we are looking at a circular clock face with charges placed at the 3 o'clock, 6 o'clock, 9 o'clock, and 12 o'clock points. The question is asking for the electric field at the center of the clock face.

To solve this problem, we can use the concept of superposition, which states that the total electric field at any point is the vector sum of the electric fields created by each individual charge. In this case, the electric field at the center of the clock face will be the sum of the electric fields created by the 3nC, 6nC, 9nC, and 12nC charges.

First, we need to calculate the distance between each charge and the center of the clock face. Since the radius of the clock face is 50cm, the distance from the center to the 3 o'clock point is also 50cm. Similarly, the distance from the center to the 6 o'clock point is 50cm, the distance to the 9 o'clock point is 50cm, and the distance to the 12 o'clock point is 50cm.

Next, we can use the formula for electric field, E = kQ/r^2, where k is the Coulomb's constant, Q is the charge, and r is the distance between the charge and the point where we want to calculate the electric field. Plugging in the values for each charge, we get:

E(3 o'clock) = (9x10^9 Nm^2/C^2)(3x10^-9 C)/(0.5m)^2 = 108 N/C
E(6 o'clock) = (9x10^9 Nm^2/C^2)(6x10^-9 C)/(0.5m)^2 = 432 N/C
E(9 o'clock) = (9x10^9 Nm^2/C^2)(9x10^-9 C)/(0.5m)^2 = 243 N/C
E(12 o'clock) = (9x10^9 Nm^2/C^2)(12x10^-9 C)/(0.5m)^2 = 432 N/C

To find the total electric field at the center of the clock face, we add these values together