Four identical metal objects carry the following charges: +17.52, ‒2.28, ‒9.77, and+19.34 C. The objects are brought simultaneously into contact, so that each touches theothers, and then they are separated.(i) Calculate the final charge on each object. [2 marks](ii) If one of the charges is then isolated from the others, determine the electric fieldstrength at a distance of 0.4 m from it, by making use of a 0.1 C test charge locatedat that point.

(i) When the objects are brought into contact, the charges will redistribute evenly among them because they are identical. To find the final charge on each object, we add up all the charges and divide by the number of objects.

The total charge is +17.52 - 2.28 - 9.77 + 19.34 = 24.81 µC.

Since there are 4 objects, the charge on each object after redistribution will be 24.81/4 = 6.2025 µC.

(ii) The electric field strength E at a distance r from a point charge q is given by Coulomb's law:

E = kq/r^2

where k is Coulomb's constant, approximately 8.99 x 10^9 N m^2/C^2.

If we isolate one of the charges (which we've found to be 6.2025 µC or 6.2025 x 10^-6 C), and want to find the electric field strength at a distance of 0.4 m from it, we substitute these values into the equation:

E = (8.99 x 10^9 N m^2/C^2) * (6.2025 x 10^-6 C) / (0.4 m)^2

This will give us the electric field strength in N/C.

To find the force on a 0.1 µC test charge located at that point, we use the equation F = Eq, where F is the force, E is the electric field strength, and q is the charge. Substituting the values we have:

F = E * 0.1 x 10^-6 C

This will give us the force on the test charge in Newtons.